Applications of Derivatives

Maxima and Minima

Exercise. A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Find the smallest area enclosed by the wire.

The length of wire is 1. Let x be the part that is bent into a circle x = 2πr, then 4s = 1-x where s is the square’s side ⇒ s = $\frac{1-x}{4}$. Besides, x = 2πr ⇒ r = $\frac{x}{2π}$. The diagram is shown in Figure 1.a.  

$A = π\frac{x^{2}}{4π^{2}}+\frac{(1-x)^{2}}{4^{2}} = \frac{4x^{2}}{4^{2}π}+\frac{π(1-x)^{2}}{4^{2}π} = \frac{4x^{2}+π(1-x)^{2}}{4^{2}π}$

$A = \frac{4x^{2}+π(1-x)^{2}}{4^{2}π} = \frac{4x^{2}+ π -2πx +πx^{2}}{4^{2}π} = \frac{(π+4)x^{2} -2πx + π}{4^{2}π}$

$A’=\frac{1}{4^{2}π}(2(π+4)x -2π)=\frac{1}{8π}((π+4)x -π)$. A’=0 ⇒ x=^π⁄_π+4

• (0, ^π⁄_π+4), A’<0 ⇒ A is decreasing.
• (^π⁄_π+4, 1), A’>0 ⇒ A is increasing. Therefore, x = ^π⁄_π+4 is a minimum.

$\frac{s}{r} = \frac{\frac{1-x}{4}}{\frac{x}{2π}} = \frac{(1-x)2π}{4x} = \frac{(1-x)π}{2x}= \frac{(1-\frac{π}{π+4})π}{2\frac{π}{π+4}} = \frac{(\frac{π+4-π}{π+4})π}{2\frac{π}{π+4}}=\frac{1}{2}$

In other words, if the radius of the circle is half the side of the square, the sum of both areas is the least.

Exercise. Find the largest area enclosed by the wire when each piece encloses a square.

The length of wire is 1, and therefore their sides are: $\frac{x}{4}, \frac{1-x}{4}$. The diagram is shown in Figure 1.a.  

A = $(\frac{x}{4})^{2}+(\frac{1-x}{4})^{2} = \frac{2x^{2}-2x+1}{16}$

A’ = $\frac{4x-2}{16}$. A’=0 ⇒ x=¹⁄₂.

• (0, ¹⁄₂), A’<0 ⇒ A is decreasing.
• (¹⁄₂, 1), A’>0 ⇒ A is increasing. Therefore, x = ¹⁄₂ is a minimum. The largest area is in x=0 or 1, that is, a single square (the problem is quite misleading) and its area is ¹⁄₁₆.

Exercise: A box with a square base and open top must have a volume of 42,592cm³. Find the dimensions of the box that minimize the amount of material used.

V = x²y = 42,592 ⇒ y = $\frac{V}{x^{2}} = \frac{42,592}{x^{2}}$ The diagram is shown in Figure 1.c.  

A = x²+4xy = x²+4^42,592⁄_x = x²+^170,368⁄_x

A’ = 2x -^170,368⁄_x². A’ = 0 ⇒ 0 = $\frac{2x^{3}-170,368}{x^{2}}$ ⇒ $x = \sqrt[3]{\frac{170,368}{2}}=44$

• (0, 44), A’<0 ⇒ A is decreasing.
• (44, ∞), A’>0 ⇒ A is increasing. x=44 is a minimum ⇒ y = 22 ⇒ Dimensions: 44cm x 44 cm x 22 cm. ⇒ A = 5,808 cm³.

A’’ = 2 + ^2*170,368⁄_x³ >0 ⇒ Concave upward.

Exercise: A box with a square base and open top must have a fixed volume. Find the dimensions of the box that minimize the amount of material used.

V = x²y (fixed) ⇒ y = $\frac{V}{x^{2}}$

A = x²+4xy = x²+4^V⁄_x

A’ = 2x -4^V⁄_x². A’ = 0 ⇒ 0 = $\frac{2x^{3}-4V}{x^{2}}$ ⇒ $x = \sqrt[3]{2V}$

• (0, $\sqrt[3]{2V}$), A’<0 ⇒ A is decreasing.
• ($\sqrt[3]{2V}$, ∞), A’>0 ⇒ A is increasing. Therefore, x = $\sqrt[3]{2V}$ is a minimum, y = $2^{\frac{-2}{3}}V^{\frac{1}{3}}$

Futhermore, $\frac{x}{y}=\frac{2^{\frac{1}{3}}V^{\frac{1}{3}}}{2^{\frac{-2}{3}}V^{\frac{1}{3}}} = 2.$

Related rates

A car and a lorry are approaching an intersection on roads that are perpendicular to each other. The car is 0.8 miles and traveling at 60mph. The lorry is 0.6 miles from the intersection and traveling at 30mph. Find the rate of change of the distance between the two cars.

Goal: $\frac{ds}{dt}$?

$\frac{dy}{dt}=-60mph, \frac{dx}{dt}=-30mph $.

The diagram is shown in Figure 1.e.  

$x^{2} + y^{2}=s^{2} ⇒ 0.6^{2} + 0.8^{2}=s^{2} ⇒ s=1$

$x^{2} + y^{2}=s^{2} ⇒ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=2s\frac{ds}{dt} ⇒ 2·0.6·(-30)+2·0.8·(-60)=2s\frac{ds}{dt} ⇒ \frac{ds}{dt}=-66mph.$

Consider a conical tank. Its radius at the top is 4 feet, and its 10 feet high. Its being filled with water at the rate of 2 cubic feet per minute. How fast is the water level rising when it is 5 feet high?

Two triangles are similar when they have equal angles and proportional sides, and therefore $\frac{r}{h} = \frac{4}{10}$ ⇒ r = $\frac{2h}{5}$. The diagram is shown in Figure 1.f.  

$V = \frac{1}{3}·base·height=\frac{1}{3}·\pi·r^{2}·h = \frac{4\pi}{75}·h^{3}$

$\frac{dV}{dt}=2$

Goal: h = 5, $\frac{dh}{dt}?$

$\frac{dV}{dt}=2=\frac{4\pi}{75}·3h^{2}·\frac{dh}{dt} = \frac{4\pi}{25}·5^{2}·\frac{dh}{dt} ⇒ \frac{dh}{dt} = \frac{1}{2\pi}$ft/sec.

A rocket is launched vertically upward from a point 2 miles west of an observer on the ground. What is the speed of the rocket when the angle of elevation of the observer’s line of sight to the rocket is 50° and is increasing at 5° per second?

We know that tan(θ)=^h⁄₂ and $\frac{dθ}{dt} = 5° per second$

Goal: $\frac{dh}{dt}~ when~ θ=50°$. The diagram is shown in Figure 2.c.  

$tan(θ)=\frac{h}{2}⇒sec^{2}θ\frac{dθ}{dt} = \frac{1}{2}\frac{dh}{dt}⇒\frac{dh}{dt}=2sec^{2}(50°·\frac{pi}{180})·5°·\frac{pi}{180} ≈ 0.422mps ≈ 1520.7mph$ by 0.422·3.600

Police are 30 feet from the side of the road. Their radar sees your car approaching at 80 feet per second when your car is 50 feet away from the radar gun. The speed limit is 65 miles per hour. Are you speeding?

We know that x²+30²=D² (⇒ x = 40) and $\frac{dD}{dt}=-80.$ Goal: $\frac{dx}{dt}?$

The diagram is shown in Figure 2.a.  

$2x\frac{dx}{dt} = 2D\frac{dD}{dt}$

Now, we can insert the values in the equation, $2x\frac{dx}{dt} = 2D\frac{dD}{dt} ⇒ 2·40·\frac{dx}{dt} = 2·50·(-80) ⇒ \frac{dx}{dt} = -100 ft/sec.$

|-100 ft/sec| > 95 ft/sec = 65mph. Yes, you are speeding, but not for much.

An airplane is flying horizontally at 480mph, 3miles above the ground when it passes a boy on the ground. How fast is the distance between the boy and the plane increasing half a minute later?

We know that 3² + b² = c², $\frac{db}{dt}=480$

480 miles ———- 60*60 seconds
b miles ———— 30 seconds

b = $\frac{480·30}{60·60} = 4 miles$ ⇒ c = 5.

The diagram is shown in Figure 2.b.  

2b$\frac{db}{dt} = 2c\frac{dc}{dt}$. Now, we can insert the values in the equation, $2·4·480 = 2·5·\frac{dc}{dt}⇒~ \frac{dc}{dt}=384mph$

Newton’s method

It is a root-finding algorithm which produces successively better approximations to the root or zeroes of a real-valued function. The idea is to start with an initial guess, e.g., x₀ = 2. Then, we approximate the function by its tangent line, and calculate the x-intercept of this tangent line. This x-intercept is our new guess, say x₁, and it will typically be a better approximation to the original function's root than the first guess. 2.d.  

The tangent line’s formula is: $y-y_0=m(x-x_0)$

x1 is the x-intercept, so $0-y_0=m(x_1-x_0) ⇒ x_1 = x_0 -\frac{y_0}{m} = x_0 -\frac{f(x_0)}{f’(x_0)}$

The process is repeated as: $x_{n+1} = x_n - \frac{f(x_n)}{f’(x_n)}$

Example. Let’s approximate $\sqrt[7]{1000}$. It is the same as trying to find the root of f(x)=x⁷ - 1000.

A reasonably good guess is x₀=3. f’(x)=7x⁶, $x_1 = x_0 -\frac{f(x_0)}{f’(x_0)} = 3 - \frac{3^{7}-1000}{7·3^{6}}≈2.76739173$

$x_2 = x_1 -\frac{f(x_1)}{f’(x_1)} = 2.76739173 - \frac{2.76739173^{7}-1000}{7·2.76739173^{6}}≈2.69008741$

$x_3 = x_2 -\frac{f(x_2)}{f’(x_2)} = 2.69008741 - \frac{2.69008741^{7}-1000}{7·2.69008741^{6}}≈2.68275645$. Google’s search engine returns 2.68269579528 ($\sqrt[7]{1000}$).