Calculus 1 - Applications of Derivative

Contents: Limit of a Function Derivative of a Function Applications of Derivative Global and Local Extrema of a Function Increasing and Decreasing of a Function Sample Problems Function of Two or More Variables

Global and Local Extrema of a Function

The function f:

• has an absolute maximum at c if f(x) ≤ f(c) for all x in the f's domain. f(c) is f's maximum value on its domain.
• has an absolute minimum at c if f(x) ≥ f(c) for all x in the f's domain. f(c) is f's minimum value on its domain.
• has a relative maximum at c if f(x) ≥ f(c) for all x in some open interval that contains c.
• has a relative maximum at c if f(x) ≤ f(c) for all x in some open interval that contains c.

A critical number of a function f, is any number c in f's domain such that f'(c) = 0 or f'(c) doesn't exist.

Suppose f has a continuous second derivative on the interval (a,b) that contains the critical number c:

• If f''(c) < 0, then f has a relative maximum at c.
• If f''(c) > 0, then f has a relative minimum at c.

Fermat's Theorem If f has a relative extremum at c, then either f'(c) = 0 or f'(c) doesn't exist.

Extreme Value Theorem If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) for some number c in [a,b] and an absolute minimum value f(d) for some number d in [a,b].

Rolle's Theorem Let f be a continuous function on [a,b] and differentiable on (a,b). If f(a) = f(b), then there exists one number c in (a,b) such that f'(c) = 0.

Mean Value Theorem Let f be a continuous function on [a,b] and differentiable on (a,b). If f(a) = f(b), then there exists one number c in (a,b) such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ Which means that: if f'(c) = 0 for all x in interval (a,b) then f is constant on (a,b). if f'(x) = g'(x) for all x in interval (a,b) then f and g differ by a consant on (a,b). Or in other words, there exists a constant c such that f(x) = g(x) + c for all x in (a,b).

Increasing and Decreasing of a Function

The function f is:

• increasing on an interval, if for every pair of numbers in that interval such that: a < b implies that f(a) < f(b)
• decreasing on an interval, if for every pair of numbers in that interval such that: a < b implies that f(a) > f(b).
• monotonic on an interval, if it is either increasing, or decreasing.

Suppose f is differentiable on an open interval (a,b).

• If f'(x) > 0 for all x in (a,b), then f is increasing on (a,b).
• If f'(x) < 0 for all x in (a,b), then f is decreasing on (a,b).
• If f'(x) = 0 for all x in (a,b), then f is constant on (a,b).

Concavity of a Function
Suppose f is differentiable on an open interval I. When the graph f is drawn:

• f'(x) > 0 indicates that f is increasing.
• f'(x) < 0 indicates that f is decreasing.
• f''(x) > 0 indicates that f is concave upwards.
• f''(x) < 0 indicates that f is concave downwards.

Let f be a function continuous on an open interval containing c and there exists a tangent line at the point P(c,f(c)). If the graph changes from concave upwards to concave downwards (or vice versa), then the point P is called the Inflection Point.

Sample Problems

Problem 1: Finding Tangent Lines
A-C: Given a curve, and a point on the said curve, find the equation of the tangent line, and the normal line.
Normal line at a point is a line that is perpendicular to the tangent line at that point.
D-E: Given two curves, show that these curves are orthogonal.
Two curves are said to be orthogonal, if their tangent lines are perpendicular at each point of intersection of the curves.

1. x⁴-2xy+y⁴=0; (1,1)
2. y=cos(xy); (0,1)
3. log₃(xy)=y; (3,1)
4. 
5. 

General step for A-C:

• Step 1 Start by implicit differentiation, and find dy/dx.
• Step 2 Substitute the point to dy/dx, so that the gradient of tangent line is obtained.
• Step 3 Use the gradient to find the tangent line equation.
• Step 4 Use the gradient to find the line equation that's perpendicular to the tangent line.

Sub-Problem A
$\begin{align*} x^4-2xy+y^4 &= 0\\ \frac{d}{dx}(x^4-2xy+y^4) &= 0\\ \left (4x^3 -2x\frac{dy}{dx} - 2y + 4y^3 \frac{dy}{dx} \right ) &= 0\\ 4x^3-2y&=\frac{dy}{dx}(2x-4y^3)\\ \frac{dy}{dx}&=\frac{4x^3-2y}{2x-4y^3}\\ \\ \frac{dy}{dx}&=\frac{4-2}{2-4}\\ &=-1 \\ (y-y_o)&=m(x-x_o)\\ y-1&=-(x-1)\\ y&=-x+2\\ \\ (y-y_o)&=m(x-x_o)\\ y-1&=(x-1)\\ y&=x\\ \end{align*}$

In conclusion:

• The tangent line: y=-x+2.
• The normal line: y=x.

Sub-Problem B
$\begin{align*} y&=\cos{xy}\\ \frac{dy}{dx}&= -\left ( y + x\frac{dy}{dx} \right ) \sin{xy}\\ \frac{dy}{dx}&= -y \sin{xy} -x\sin{xy}\frac{dy}{dx}\\ \frac{dy}{dx}(1+x\sin{xy})&=-y\sin{xy}\\ \frac{dy}{dx}&=\frac{-y\sin{xy}}{1+x\sin{xy}}\\ \\ \frac{dy}{dx}&=\frac{-y\sin{0}}{1+x\sin{0}}\\ &=0 \\ (y-y_o)&=m(x-x_o)\\ y-1&=0\\ y&=1\\ \end{align*}$

Since the tangent line (y=1) is a horizontal line, then the normal line would be a vertical line that passes through the point (0,1).

In conclusion:

• The tangent line: y=1.
• The normal line: x=0.

Sub-Problem C
$\begin{align*} \log_3{xy}&=y\\ \frac{x\frac{dy}{dx}+y}{xy\ln{3}} &= \frac{dy}{dx}\\ \frac{dy}{dx} \frac{1}{y\ln{3}} + \frac{1}{x\ln{3}} &= \frac{dy}{dx}\\ \frac{1}{x\ln{3}} &= \frac{dy}{dx} - \frac{dy}{dx} \frac{1}{y\ln{3}}\\ \frac{1}{x\ln{3}} &= \frac{dy}{dx} \left (1- \frac{1}{y\ln{3}} \right)\\ \frac{1}{x\ln{3}} &= \frac{dy}{dx} \left (\frac{y\ln{3}-1}{y\ln{3}} \right)\\ \frac{dy}{dx} &= \frac{y}{x(y\ln{3}-1)}\\ \\ \frac{dy}{dx} &= \frac{y}{x(y\ln{3}-1)}\\ &= \frac{1}{3(\ln{3}-1}\\ &= \frac{1}{3\ln{3}-3} \\ (y-1)&=m(x-3)\\ y-1&=\frac{1}{3\ln{3}-3}(x-3)\\ y&=\frac{1}{3\ln{3}-3}(x-3)+\frac{3\ln{3}-3}{3\ln{3}-3}\\ &=\frac{x}{3\ln{3}-3}+\frac{\ln{3}-2}{\ln{3}-1}\\ \\ (y-1)&=m(x-3)\\ y-1&=(3-3\ln{3})(x-3)\\ y&=(3-3\ln{3})(x-3)+1\\ &=(3x-9-3x\ln{3}+9\ln{3})+1\\ &=(3-3\ln{3})x + (9\ln{3}-8) \\ \end{align*}$

In conclusion:

• The tangent line: $y=\frac{x}{3\ln{3}-3}+\frac{\ln{3}-2}{\ln{3}-1}$
• The normal line: $y=(3-3\ln{3})x + (9\ln{3}-8)$

General step for D-E:

• Step 1 Find the intersection points of the given curves.
• Step 2 Find the gradient of the tangent line, at the intersection points.
• Step 3 Check if both gradient fulfill m₁m₂=-1.

Sub-Problem D
From the graph, it can be seen that the x-coordinate of the intersection point is 0. Therefore, we can just substitute x=0 to any of the given curve to obtain the intersection point.
$\begin{align*} 2y-2x&=\pi\\ 2y&=\pi\\ y&=\frac{\pi}{2} \end{align*}$

Next, we find the gradient of the tangent lines of both curves during that point:
$\begin{align*} 2y-2x&=\pi\\ \frac{d}{dx}(2y-2x)&=0\\ 2\frac{dy}{dx}-2&=\\ \frac{dy}{dx}&=1\\ m_{1}&=1\\ \\ x&=\cos{y}\\ \frac{d}{dx}(x)&=\frac{d}{dx}\cos{y}\\ 1&=-\frac{dy}{dx}\sin{y}\\ \frac{dy}{dx}&=-\frac{1}{\sin{y}}\\ \frac{dy}{dx}&=-\frac{1}{\sin{\frac{\pi}{2}}}\\ \frac{dy}{dx}&=-1\\ m_{2}&=-1\\ \\ m_{1}m_{2}&=-1\\ (1)(-1)&=-1 \end{align*}$

The tangent lines of both curves at the intersection points are orthogonal to the other. Therefore, both curves are orthogonal.


Sub-Problem E
From the graph, it can be seen that the x-coordinate of the intersection points are -1 and 1. Therefore, we can just substitute x=-1 and x=1 to any of the given curve to obtain the intersection points.
$\begin{align*} x^2+y^2&=2\\ (-1)^2+y^2&=2\\ y^2&=1\\ y&=1\\ \end{align*}$

Next, we find the gradient of the tangent lines of both curves during both points (-1,1) and (1,1). First, the point (-1,1):
$\begin{align*} 2y&=x^2-1\\ \frac{d}{dx}(2y)&=\frac{d}{dx}(x^2-1)\\ 2\frac{dy}{dx} &= 2x\\ \frac{dy}{dx} &= x\\ m_{1}&=-1\\ \\ x^2+y^2&=2\\ \frac{d}{dx}(x^2+y^2)&=0\\ 2x+2y\frac{dy}{dx}&=0\\ \frac{dy}{dx}&=-x\\ m_{2}&=1\\ \\ m_{1}m_{2}&=-1\\ (-1)(1)&=-1\\ \\ \end{align*}$

The tangent lines of both curves at (-1,1) are orthogonal. Now, checking at the point (1,1). Since we have found the formula for dy/dx, we can just substitute the variables directly:
$\begin{align*} \frac{dy}{dx} &= x\\ m_{1}&=-1\\ \\ \frac{dy}{dx}&=-x\\ m_{2}&=-1\\ \\ m_{1}m_{2}&=-1\\ (1)(-1)&=-1\\ \\ \end{align*}$

The tangent lines of both curves at (1,1) are orthogonal. Since the tangent lines are perpendicular at each point of the intersection of the curves, both curves are orthogonal.

Problem 2: Related Rates

1. A 17-ft ladder leaning against a wall begins to slide. How fast is the top of the ladder sliding down the wall at the instant of time when the bottom of the ladder is 8 ft from the wall and sliding away from the wall at the rate of 5 ft/sec?
2. A coffee pot that has the shape of a circular cylinder of radius 8 cm is being filled with water flowing at a constant rate. At what rate is the water flowing into the coffee pot when the water level is rising at the rate of 0.8 cm/sec?
3. Two ships, A and C leave the same port at noon. Ship A moves exactly north at 16 km/h, and ship B moves east at 12 km/h. How fast is the distance between them changing at 1 P.M.?
4. The equation V = IR shows the relation between voltage V in volts (V), the current I in amperes (A) and the resistance R in ohms (Ω). When V = 24 and I = 4, V is increasing at the rate of 2 V/sec, and I is increasing at the rate of 4 A/sec, how fast is the resistance changing? Is it increasing or decreasing?

Sub-Problem A
Every point on the ladder (like the bottom of the ladder, and the top of the ladder) will fulfill x² + y² = 17². By the phytagorean theorem, we know that the wall's height is 15. From the question we know that for x = 8, then dx/dt = 5. Now, we are trying to find dy/dt:

$\begin{align*} x^2+y^2&=17^2\\ \frac{d}{dt}(x^2+y^2)&=\frac{d}{dt}17^{2}\\ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} &= 0\\ 2(8)(5) + 2(16)\frac{dy}{dt}&= 0\\ 80 + 32\frac{dy}{dt}&= 0\\ \frac{dy}{dt}&=-\frac{5}{2} \end{align*}$

Sub-Problem B
From the question it is known that dh/dt = 4/5. The following is the formula of the volume of a cylinder:
$\begin{align*} V&=\pi r^2 h\\ \frac{d}{dt}(V)&=\frac{d}{dt}(\pi r^2 h)\\ \frac{dV}{dt} &=\pi r^2 \frac{dh}{dt}\\ \frac{dV}{dt} &=\pi(64)\left ( \frac{4}{5} \right ) \\ \frac{dV}{dt} &=\frac{256}{5}\pi\\ \end{align*}$

So, the water is flowing into the coffee pot at the rate of 51.2π cm³/sec.

Sub-Problem C
At 1 P.M, ship A would be 16km from the port and ship B would be 12km from the port. Which means, the distance between the two ships (at 1 P.M.) can be found through the phytagorean theorem:

$\begin{align*} c^2&=a^2+b^2\\ c^2&=256+144\\ c&=20\\ \end{align*}$

So, at 1 P.M, the distance between both ships are 20 km. Next, we know that da/dt and db/dt are the velocity of both ships that have been mentioned in the question. So, dc/dt is the rate of the distance of both ships changing:
$\begin{align*} c^2&=a^2+b^2\\ \frac{d}{dt} (c^2)&=\frac{d}{dt} (a^2+b^2)\\ 2c \frac{dc}{dt} &= 2a \frac{da}{dt} + 2b \frac{db}{dt}\\ 2(20) \frac{dc}{dt} &= 2(16)(16) + 2(12)(12)\\ 40 \frac{dc}{dt} &= 512+288\\ \frac{dc}{dt} &= 20\\ \end{align*}$

So, the rate of the distance of both ships changing is 20 km/h.

Sub-Problem D
Differentiate V = IR with respect to t (time). Then, substitute the variables mentioned in the problem text.

$\begin{align*} V&=IR\\ \frac{d}{dt}(V)&=\frac{d}{dt}(IR)\\ \frac{dV}{dt} &= I\frac{dR}{dt} + R\frac{dI}{dt}\\ 2 &= 4 \frac{dR}{dt} + \frac{V}{I} (4)\\ 2 &= 4 \frac{dR}{dt} + 24\\ -22 &= 4 \frac{dR}{dt}\\ \frac{dR}{dt} &= -\frac{11}{2} \end{align*}$

So, the resistance is changing at the rate of -5,5 Ω/sec. Since it's negative, the resistance is decreasing.

Problem 3: Finding Extrema
Given f(x) and the interval, find the absolute maximum or minimum of the function at the given interval if it exists.
$\begin{align*} \text{A. }&f(x)= 2\sin{x} - x\\ &[0,\frac{\pi}{2}]\\ \text{B. }&f(x)= (x^2-4)\sqrt[3]{x^2}\\ &[-1,2]\\ \text{C. }&f(x)= x^2 - x -2\\ &[0,2]\\ \text{D. }&f(x)= \frac{x}{x^2+1}\\ &[-1,2]\\ \text{E. }&f(x)= \ln{x^{2}+4x+5}\\ &[-4,-2] \end{align*}$

General step:

• Step 1 Find all critical points.
• Step 2 Consider the endpoints of the interval.
• Step 3 Check all critical points that is in the given interval, and also the endpoints.

Sub-Problem A
$\begin{align*} f(x)&= 2\sin{x} - x\\ f'(x)&= 2\cos{x} - 1\\ \cos{x}&=\frac{1}{2}\\ \\ x&=\frac{\pi}{3}\\ \end{align*}$

So, with the interval [0, 90°] we have:

x	f(x)	Max/Min?
0	0	Min
π/3	√3 - π/3	Max
π/2	2 - π/2	Neither

Sub-Problem B
$\begin{align*} f(x)&= (x^2-4)\sqrt[3]{x^2}\\ f'(x)&= (2x)\sqrt[3]{x^2} + \frac{2}{3\sqrt[3]{x}}(x^2-4)\\ 0&= (2x)\sqrt[3]{x^2} + \frac{2}{3\sqrt[3]{x}}(x^2-4)\\ -\frac{2}{3\sqrt[3]{x}}(x^2-4)&=(2x)\sqrt[3]{x^2}\\ 2(x^2-4)&=-6x^2\\ x^2-4&=-3x^2\\ x^2&=1\\ \end{align*}$

So, with the interval [-1, 2] we have:

x	f(x)	Max/Min?
-1	-3	Min
0	0	Max
1	-3	Min
2	0	Max

Sub-Problem C
$\begin{align*} f(x)&= x^2-x-2\\ f'(x)&= 2x-1\\ x&=\frac{1}{2} \end{align*}$

So, with the interval [0, 2] we have:

x	f(x)	Max/Min?
0	-2	Neither
0.5	-2.25	Min
2	0	Max

Sub-Problem D
$\begin{align*} f(x)&= \frac{x}{x^2+1}\\ f'(x)&= \frac{1-x^2}{(x^2+1)^2}\\ 0&=1-x^2\\ x&=\pm 1\\ \end{align*}$

So, with the interval [-1, 2] we have:

x	f(x)	Max/Min?
-1	-0.5	Min
1	0.5	Max
2	0.4	Neither

Sub-Problem E
$\begin{align*} f(x)&= \ln\left(x^{2}+4x+5\right)\\ f'(x)&= \frac{2x+4}{x^{2}+4x+5}\\ 0&=2x+4\\ x&=-2\\ \end{align*}$

So, with the interval [-4, -2] we have:

x	f(x)	Max/Min?
-4	ln(5)	Max
-2	ln(1)	Min

Problem 4: Finding Intervals
Given f(x), find the intervals in which f is increasing and/or decreasing, and the critical numbers if it exists.
$\begin{align*} \text{A. }&f(x)= x^3-3x+4\\ \text{B. }&f(x)= \ln{x^2}\\ \text{C. }&f(x)= x+\frac{1}{x}\\ \text{D. }&f(x)= \frac{x+2}{\sqrt{x^2+4}}\\ \text{E. }&f(x)= \left|x^{3}-3x^{2}+4\right|\\ \end{align*}$

Sub-Problem A
$\begin{align*} f(x)&= x^3-3x+4\\ f'(x)&=3x^2-3\\ 0&=3(x^2-1)\\ \\ x&=1\\ x&=-1 \end{align*}$

Now we know that the critical numbers are 1 and -1. Next, we find the intervals. Let us test three values a, b and c for x, such that a < -1, -1 < b < 1, then c > 1. For example, a = -2, b = 0, and c = 2:
$\begin{align*} f'(x)&=3x^2-3\\ f'(-2)&=3(-2)^2-3\\ &=9 \end{align*}$

Since f'(-2) > 0, it can be concluded that the function is an increasing function for x < -1. Next:
$\begin{align*} f'(x)&=3x^2-3\\ f'(0)&=3(0)^2-3\\ &=-3 \end{align*}$

Since f'(0) < 0, it can be concluded that the function is a decreasing function for -1 < x < 1. Next:
$\begin{align*} f'(x)&=3x^2-3\\ f'(2)&=3(2)^2-3\\ &=9 \end{align*}$

Since f'(2) > 0, it can be concluded that the function is an increasing function for x > 1.

In conclusion, the given function f:

• has the critical numbers -1 and 1
• is increasing on: -∞ < x < -1 or 1 < x < ∞
• is decreasing on: -1 < x < 1

Sub-Problem B
$\begin{align*} f(x)&= \ln{x^2}\\ f'(x)&= \frac{2}{x} \\ \end{align*}$

Though x=0 for f'(x) will make f'(x) undefined, x=0 is not in f's domain. The domain of f is x < 0 or x > 0. So, there are no critical numbers.

By observing f'(x), it is clear to see that x > 0 implies f'(x) > 0, and x < 0 implies f'(x) < 0.

In conclusion, the given function f:

• has no critical numbers.
• is increasing on: 0 < x < ∞
• is decreasing on: -∞ < x < 0

Sub-Problem C
$\begin{align*} f(x)&= x+\frac{1}{x}\\ f'(x)&= 1 - \frac{1}{x^2} \\ 0 &= \frac{x^2-1}{x^2}\\ x&=1\\ x&=-1\\ \end{align*}$

Though x=0 for f'(x) will make f'(x) undefined, x=0 is not in f's domain. The domain of f is x < 0 or x > 0. So, the only critical numbers are 1 and -1. Next, we find the intervals.

$\begin{align*} f'(x)&=\frac{x^2-1}{x^2} \end{align*}$

By observing f'(x), it is clear to see that any values of x (x ≠ 0), the denominator will always be a positive number. The numerator part is the same as f'(x) in Sub-Problem A.

In conclusion, the given function f:

• has the critical numbers -1 and 1.
• is increasing on: -∞ < x < -1 or 1 < x < ∞
• is decreasing on: -1 < x < 1 and x ≠ 0

Sub-Problem D
$\begin{align*} f(x)&= \frac{x+2}{\sqrt{(x^2+4)}}\\ f'(x)&= \frac{4-2x}{\sqrt{\left(x^{2}+4\right)^{3}}}\\ 0 &= \frac{4-2x}{\sqrt{\left(x^{2}+4\right)^{3}}}\\ &= 4-2x\\ x&=2\\ \end{align*}$

So, the critical number is 2. From the simplified form above (4-2x), it is clear to see that x < 2 implies f'(x) > 0, and vice versa.

In conclusion, the given function f:

• has the critical number 2.
• is increasing on: -∞ < x < 2
• is decreasing on: 2 < x < ∞

Sub-Problem E
Let x³-3x²+4:
$\begin{align*} f(x)&= \left|x^{3}-3x^{2}+4\right|\\ &= \left | u right |\\ f'(x)&= \frac{u}{|u|}u'\\ &= \frac{x^{3}-3x^{2}+4}{|x^{3}-3x^{2}+4|} (3x^{2}-6x) \\ 0&=(x^{3}-3x^{2}+4)(3x^{2}-6x)\\ &=(x-2)(x-2)(x+1)3x(x-2) \end{align*}$

We must inspect x³-3x²+4 and 3x²-6x, since they determine whether f'(x) is greater than 0 or not.

x³-3x²+4 can be factorized into (x-2)(x-2)(x+1). From that, we could determine that x³-3x²+4 is greater than 0 on (-1,2) and (2,∞).

3x²-6x can be factorized into 3x(x-2). From that, we could determine that 3x²-6x is greater than 0 on (-∞,0) and (2,∞).

Since both x³-3x²+4 and 3x²-6x are greater than 0 on (2,∞), we can conclude that f(x) is increasing on (2,∞).

Since both x³-3x²+4 and 3x²-6x are greater than 0 on (-1,0), we can conclude that f(x) is increasing on (-1,0).

Since only x³-3x²+4 is positive on (0,2), and 3x²-6x is negative during that interval, we can conclude that f(x) is decreasing on (0,2).

Since only 3x²-6x is positive on (-∞,-1), and x³-3x²+4 is negative during that interval, we can conclude that f(x) is decreasing on (-∞,-1). In conclusion, the given function f:

• has the critical numbers -1, 0 and 2.
• is increasing on: (-1,0) and (2,∞)
• is decreasing on: (-∞,-1) and (0,2)

Problem 5: Finding Intervals
Given f(x), find the intervals of the concavity of the given function, and the inflection points.
$\begin{align*} \text{A. }&f(x)= \frac{1}{3}x^{3}+3x^{2}+5\\ \text{B. }&f(x)= \sin{x}\\ \text{C. }&f(x)= \ln{x}\\ \text{D. }&f(x)= \frac{x}{x-1}\\ \text{E. }&f(x)= \frac{x^{2}+4x+4}{x^{2}+4x+3}\\ \end{align*}$

General step:

• Step 1 Find the number c, such that f''(c) = 0.
• Step 2 Recall that when f''(x) > 0, the function will be concave upwards and when f''(x) < 0, the function will be concave downwards. Identify the concavity of the function with that information.

Sub-Problem A
$\begin{align*} f(x)&= \frac{1}{3}x^{3}+3x^{2}+5\\ f'(x)&=x^2+6x\\ f''(x)&=2x+6\\ 0&=2x+6\\ x&=-3\\ \end{align*}$

From f''(x) above, we can conclude that the function f:

• is concave upwards on (-3,∞)
• has the inflection point during x = -3.
• is concave downwards on (-∞,-3)

Sub-Problem B
$\begin{align*} f(x)&= \sin{x}\\ f'(x)&= \cos{x} \\ f''(x)&=-\sin{x}\\ 0&=-\sin{x}\\ x&=n\pi\text{, where }n\epsilon \mathbb{Z} \end{align*}$

From f''(x) above, we can conclude that the function f: (Let n be an integer)

• is concave upwards on (2πn-π,2πn)
• has the inflection point during x = nπ.
• is concave downwards on (2πn,2πn+π)

Sub-Problem C
$\begin{align*} f(x)&= \ln{(x)}\\ f'(x)&= \frac{1}{x} \\ f''(x)&= -\frac{1}{x^2}\\ \end{align*}$

From f''(x) above, we can conclude that the function f:

• will never be concave upwards.
• has no inflection point.
• is concave downwards on (0,∞)

Sub-Problem D
$\begin{align*} f(x)&= \frac{x}{x-1}\\ f'(x)&= -\frac{1}{(x-1)^2} \\ f''(x)&= \frac{2}{(x-1)^3}\\ 0&=\frac{2}{(x-1)^3}\\ \end{align*}$

From f''(x) above, we can conclude that the function f:

• is concave upwards on (1,∞)
• has no inflection point.
• is concave downwards on (-∞,1)

Sub-Problem E
$\begin{align*} f(x)&= \frac{x^{2}+4x+4}{x^{2}+4x+3}\\ f'(x)&= \frac{(2x+4)(x^{2}+4x+3)-(2x+4)(x^{2}+4x+4)}{(x^{2}+4x+3)^2} \\ &= -\frac{(2x+4)}{(x^{2}+4x+3)^2} \\ f''(x)&= \frac{-2(x^{2}+4x+3)^2 + (2x+4) 2 (2x+4) (x^{2}+4x+3)}{(x^{2}+4x+3)^4} \\ &= \frac{-2(x^{2}+4x+3) + 2(2x+4)^2 }{(x^{2}+4x+3)^3} \\ &= \frac{-2x^{2}-8x-6 + 8x^2+32x+32 }{(x^{2}+4x+3)^3} \\ &= \frac{6x^{2}+24x+26 }{(x^{2}+4x+3)^3} \\ 0&= 6x^{2}+24x+26 \\ \\ D&=\sqrt{b^2-4ac}\\ &=\sqrt{576-4(6)(26)}\\ &=\sqrt{576-624} \end{align*}$

Since the discriminant < 0, there would be no solution for f''(x) meaning that there is no inflection point.

Though finding interval for f''(x) may seem hard, for this problem, we only need to inspect the denominator part. Because, the numerator part will always be positive for all x (The fact that the lowest value possible of the numerator is 2, which can be found through simple derivation). A negative number a, when cubed (a³) stays negative. So, we only need to find interval for (x²+4x+3). Moreover, it can be factorized (x+3)(x+1) too. That simplifies things.

From f''(x) above, we can conclude that the function f:

• is concave upwards on (-∞,-3) and (-1,∞)
• has no inflection point.
• is concave downwards on (-3,-1)

Problem 6: Optimization Problems

1. Find the dimensions of the blue rectangle inscribed in the ellipse below, so that the blue rectangle has the maximum area.
   
2. Suppose there is a carton paper with the dimensions 24 m x 9 m, and an open box will be made out of it. Find the dimensions of the box with the largest volume that can be made, and what is the maximum volume possible?
3. Suppose there is a carton paper with the following shape, and will be used to form a cone. Find the θ, so that the formed cone will have maximum volume.
   
4. Find the points from the hyperbole 4x² + y² = 16 that are closest to the point (0,1).
5. Suppose there exists three points: A(-7,2), B(5,5) and C(k,0). Find the value for k, such that the total length of the line AC and BC is minimized.

Sub-Problem A
We know that:

• The equation for the ellipse shown above can be used to find a point exactly located in the circle.
• The area of the blue rectangle is equal to 2xy.

The variable y represents the rectangle's height, while the variable x represents half of the rectangle's width.
$\begin{align*} x^2+4y^2&=4\\ 4y^2&=4-x^2\\ y&= \sqrt{1-\frac{1}{4}x^2}\\ A&=2xy\\ \end{align*}$

So:
$\begin{align*} A&=2xy\\ &=2x\left (\sqrt{1-\frac{1}{4}x^2} \right )\\ A'&= \frac{2(2x-x^3)}{2\sqrt{x^2-\frac{1}{4}x^4}}\\ &= \frac{(2-x^2)}{\sqrt{1-\frac{1}{4}x^2}}\\ 0&=2-x^2\\ x&=\sqrt{2} \end{align*}$

Lastly, substitute x in the original ellipse equation:
$\begin{align*} y&= \sqrt{1-\frac{1}{4}x^2}\\ &= \sqrt{1-\frac{1}{2}}\\ &=\frac{\sqrt{2}}{2} \end{align*}$

So, in order to achieve the maximum area (which is exactly 2):

• The blue rectangle's length must be $2\sqrt{2}$,
• The blue rectangle's height must be $\frac{\sqrt{2}}{2}$

Sub-Problem B
The variable y will represent the volume, and x will represent the height of the carton box. Imagine the net of a cuboid inscribed inside a rectangle:

By looking at the visualization above, the volume of the box can be written as the following equation, and then by finding the derivative of it, the largest volume of the box can be found:
$\begin{align*} y&=(24-2x)(9-2x)x\\ &=216x-66x^2+4x^3\\ y'&=216-132x+12x^2\\ &=12(18-11x+x^2)\\ &=12(x-9)(x-2)\\ \end{align*}$

So, we have x=9 and x=2. If we look at the visualization once again, it is impossible for x=9. So, x must be 2. Substitute them into the original y:
$\begin{align*} y&=(24-2x)(9-2x)x\\ &=(20)(5)(2)\\ &=400 \end{align*}$

Therefore, in order to achieve the maximum volume of the box (which is 200m³), the dimension of the box must be 20m x 5m x 2m.

Sub-Problem C
We know that:

• Circumference of the whole original circle is just the circumference of the cone and arc formed by θ combined.
• Phytagorean theorem represents the relationship between r, t and R.
• Volume of cone is one third of the bottom circle's area times its height.

$\begin{align*} (2\pi r) + (R\theta) &= (2\pi R)\\ t^2+r^2&=R^2\\ V&=\frac{1}{3}\pi r^2 t\\ \end{align*}$

In order to achieve maximum volume:
$\begin{align*} V&=\frac{1}{3}\pi r^2 t\\ &=\frac{1}{3}\pi (R^2-t^2) t\\ &= \frac{1}{3}\pi (R^2t-t^3)\\ V'&= \frac{1}{3}\pi (R^2-3t^2)\\ 0&=\frac{1}{3}\pi (R^2-3t^2)\\ 3t^2&=R^2\\ t\sqrt{3}&=R\\ t&=\frac{\sqrt{3}}{3}R\\ \end{align*}$

Now, substitute t:
$\begin{align*} t^2+r^2&=R^2\\ r^2&=R^2-t^2\\ r&=\sqrt{R^2-t^2}\\ &=\sqrt{R^2-\left(\frac{\sqrt{3}}{3}R\right) ^2}\\ &=\sqrt{R^2-\frac{1}{3}R^2}\\ &=\frac{R\sqrt{6}}{3} \end{align*}$

Now, we use our first equation:
$\begin{align*} (2\pi r) + (R\theta) &= (2\pi R)\\ R\theta &= 2\pi(R-r)\\ &=2\pi \left ( R-\frac{R\sqrt{6}}{3} \right )\\ \theta&=2\pi \left ( \frac{3-\sqrt{6}}{3} \right )\\ \end{align*}$

If approximated, then θ so that the volume of the cone is maximum, is 66.061° degrees.

Sub-Problem D
$\begin{align*} x^2-\frac{y^2}{4}&=4\\ x^2&=4+\frac{y^2}{4}\\ x&=\sqrt{4+\frac{y^2}{4}} \end{align*}$

We use the formula when finding distance between two points. That is:
$\begin{align*} D&=\sqrt{(x-x_o)^2+(y-y_o)^2}\\ &=\sqrt{x^2+(y-1)^2}\\ &=\sqrt{4+\frac{y^2}{4} + y^2 - 2y + 1}\\ &=\sqrt{\frac{5}{4}y^2 -2y + 5}\\ D'&=\frac{\frac{5}{2}y-2}{2\sqrt{\frac{3}{4}y^2 -2y + 5}}\\ 0&=\frac{\frac{5}{2}y-2}{2\sqrt{\frac{3}{4}y^2 -2y + 5}}\\ 0&=\frac{5}{2}y-2\\ y&=\frac{4}{5}\\ \end{align*}$

That means, the distance will be minimum when y = ⁴/₅. Next, we substitute the y into the original hyperbole equation to find the point:
$\begin{align*} x&=\sqrt{4+\frac{y^2}{4}}\\ &=\sqrt{4+\frac{4}{25}}\\ &=\pm\frac{2}{5}\sqrt{26} \end{align*}$

So, the two points that are closest to (0,1) are:
$\begin{align*} &\left ( \frac{2}{5}\sqrt{26},\frac{4}{5} \right )\\ &\left ( -\frac{2}{5}\sqrt{26},\frac{4}{5} \right ) \end{align*}$

Sub-Problem E
The variable m will represent the C's horizontal distance from A:
$\begin{align*} D_{AC}&=\sqrt{(x-x_A)^2+(y-y_A)^2}\\ &=\sqrt{(m)^2+(-2)^2}\\ &=\sqrt{m^2+4}\\ \\ D_{BC}&=\sqrt{(12-m)^2+(y-y_B)^2}\\ &=\sqrt{m^2-24m+169}\\ \end{align*}$

Now, the main function we have to differentiate is the sum of the above:
$\begin{align*} D&=\sqrt{m^2+4}+\sqrt{m^2-24m+169}\\ D'&=\frac{2m}{2\sqrt{m^2+4}} + \frac{2m-24}{2\sqrt{m^2-24m+169}}\\ 0&= \frac{m}{\sqrt{m^2+4}} + \frac{m-12}{\sqrt{m^2-24m+169}}\\ \frac{12-m}{\sqrt{m^2-24m+169}} &= \frac{m}{\sqrt{m^2+4}}\\ \frac{(12-m)^2}{(12-m)^2+25} &= \frac{m^2}{m^2+4}\\ \frac{(12-m)^2+25}{(12-m)^2} &= \frac{m^2+4}{m^2}\\ 1 + \frac{25}{(12-m)^2} &= 1 + \frac{4}{m^2}\\ \frac{25}{(12-m)^2} &= \frac{4}{m^2}\\ \frac{5}{(12-m)} &= \frac{2}{m}\\ 5m &= 2(12-m)\\ 5m &= 24 - 2m\\ m &= \frac{24}{7} \end{align*}$

Lastly, since this value of "m" we've got is only the horizontal distance between C and A:


Which means, k=-7+m Therefore, the value for k such that the sum of the line AC and BC is minimum, is -²⁵/₄.