lecture 25: Optimization Problems and Related Rates

Optimization Review

Recall the strategy:

Identify the function to optimize.
Write it in terms of one variable.
Find the derivative, solve for critical points.
Test points to find max/min.
Interpret in context.

Application: Minimizing Cost

Problem:

You’re building a box with a square base and open top. It must have a volume of 32,000 cm³. What dimensions minimize the surface area?

Steps:

Let the base side = $x$ , height = $h$
Volume: $x^2 h = 32{,}000 \Rightarrow h = \frac{32{,}000}{x^2}$
Surface Area (no top): $A(x) = x^2 + 4xh = x^2 + 4x\left(\frac{32{,}000}{x^2}\right) = x^2 + \frac{128{,}000}{x}$
Derivative: $A'(x) = 2x - \frac{128{,}000}{x^2}$
Set $A'(x) = 0$ : $2x = \frac{128{,}000}{x^2} \Rightarrow 2x^3 = 128{,}000 \Rightarrow x = 40$
Then $h = \frac{32{,}000}{40^2} = 20$

Min surface area when base is 40 cm × 40 cm and height is 20 cm.

Introduction to Related Rates

Related rates describe how two or more quantities change together over time.

Key Idea:

Take the derivative with respect to time $t$ , using the chain rule.

Steps for Solving Related Rates Problems

Identify all changing quantities.
Write an equation that relates them.
Differentiate with respect to time $t$ .
Plug in known values and solve for the unknown rate.
Include units and interpret your result.

Application: Balloon Inflation

Problem:

Air is being pumped into a spherical balloon at a rate of 100 cm³/s. How fast is the radius increasing when the radius is 5 cm?

Steps:

Volume of a sphere: $V = \frac{4}{3}\pi r^3$
Differentiate: $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$
Given: $\frac{dV}{dt} = 100$ , $r = 5$
Plug in: $100 = 4\pi (25) \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318 \text{ cm/s}$

Radius increasing at ~0.318 cm/s

Application: Sliding Ladder

Problem:

A 10 ft ladder is leaning against a wall. The bottom slides away at 2 ft/s. How fast is the top sliding down when the bottom is 6 ft from the wall?

Setup:

Use Pythagorean Theorem: $x^2 + y^2 = 100$
Differentiate: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
Given: $x = 6$ , $\frac{dx}{dt} = 2$

Find $y$ : $6^2 + y^2 = 100 \Rightarrow y = 8$

Plug in:

2(6)(2) + 2(8)\frac{dy}{dt} = 0 \Rightarrow 24 + 16\frac{dy}{dt} = 0

\frac{dy}{dt} = -\frac{3}{2} = -1.5 \text{ ft/s}

Top of ladder descending at 1.5 ft/s

Exercises

1. Expanding Circle

A circle’s area increases at 10 cm $^2$ /s. How fast is the radius changing when the area is $100\pi$ cm $^2$ ?

2. Draining Tank

Water drains from a cylindrical tank at $t^2$ m $^3$ /min (= $t^2$ cubic meters per minute). What’s the rate of change of the water depth at $t=3$ if the radius is 1 m? (Volume of a cylinder is $\pi r^2 h$ )

3. Cone Volume

A conical water tank is being filled at 3 m $^3$ /min. The tank’s radius is always half its height. If the total height $h=10$ m, how fast is the water level rising when the tank is 4 m deep? (Volume of a cone is $\frac{1}{3}\pi r^2 h$ )

4. Sliding Ladder

A 13-foot ladder is leaning against a vertical wall. The top of the ladder is sliding down the wall at a rate of 1.5 ft/s. How fast is the bottom of the ladder sliding away from the wall when the top is 5 feet above the ground?

Answer key

$\frac{dA}{dt} = 10$ cm $^2$ /s, $A=\pi r^2 \implies \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{10}{2\pi \cdot 10} = \frac{1}{2\pi}$ when $A=100\pi$ cm $^2$ or $r=10$ cm.
$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{t^2}{\pi} = \frac{9}{\pi}$ when $t=3$ and $r=1$ m.
$V = \frac{\pi}{3} \cdot 5^2 \cdot 10 - \frac{\pi}{3} r^2 (10 - h) = \frac{\pi}{3} \cdot 250 - \frac{\pi}{3} (\frac{10-h}{2})^2 (10 - h)$ and given $\frac{dV}{dt} = 3$ m $^3$ /min $\implies \frac{dV}{dt} = 3 = - \frac{\pi}{3} \frac{d}{dt} [\frac{(10-h)^3}{4}] \implies -\frac{9 \times 4}{\pi} = \frac{d}{dt} [(10-h)^3] = 3(10-h)^2 \times (-1) \times \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{12}{\pi (10-h)^2} = \frac{12}{\pi \times 36} = \frac{1}{3\pi}$ when $h=4$ m.
$x^2 + y^2 = 169 \implies 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \implies \frac{dx}{dt} = - \frac{y}{x} \frac{dy}{dt} = - \frac{\sqrt{169-25}}{5} \times (-1.5) = \frac{12}{5} \times \frac{3}{2} = \frac{18}{5} = 3.6$ ft/s when $y=12$ ft.