5.10 - Introduction to Optimization Problems

What Is Optimization?

Optimize means to make the best or most effective use of a situation or resource. In calculus, that means finding:

Maximum values (largest, most, greatest)

Minimum values (smallest, least, shortest)

These show up constantly in business, engineering, economics, and physics.

The Optimization Process

This lesson covers the first three steps.

Step 1: Draw a picture (if it applies). Sketch the situation to see the known and unknown quantities.

Step 2: Write an equation to optimize. Figure out what quantity you need to max or min, then write an equation for it.

Step 3: Express it in terms of one variable. Use constraint equations to eliminate extra variables. You need one variable before you can take the derivative.

Steps 4 and 5 (taking derivatives and solving) come in the next lesson. This one is all setup.

Example 1: Product of Two Numbers

Problem: Find two positive numbers whose sum is 24 and whose product is as large as possible.

What to optimize: the product, and we want it as large as possible, so a maximum.

Step 1: No picture needed here.

Step 2: Let the two numbers be

x

and

y

P = xy

Step 3: The constraint is that the sum is 24.

x + y = 24 \quad \Rightarrow \quad y = 24 - x

Substitute:

P = x(24 - x) = 24x - x^2

Result: the equation to optimize is

P(x) = 24x - x^2

, and it’s in one variable.

Next lesson, we’d find

P'(x)

, set it equal to zero, and solve.

Example 2: Distance to a Curve

Problem: What point on the graph

y = \sqrt{x}

is closest to the point

(6, 0)

5.10 - Introduction to Optimization Problems

What to optimize: distance, and we want it smallest, so a minimum.

Step 2: Use the distance formula.

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The two points are

(x, \sqrt{x})

on the curve and

(6, 0)

d = \sqrt{(x - 6)^2 + (\sqrt{x} - 0)^2}

Step 3:

d = \sqrt{(x - 6)^2 + x}

That’s already one variable. The curve equation

y = \sqrt{x}

gave us the y-coordinate in terms of

x

for free.

Result:

d(x) = \sqrt{(x-6)^2 + x}

Handy trick: it’s often easier to minimize

d^2

instead of

d

, since that kills the square root and they have the same critical points.

Example 3: Box with Maximum Volume

Problem: A rectangular piece of cardboard measuring 20 inches by 32 inches has squares cut from each corner. The sides fold up to make an open-top box. What size squares maximize the volume?

What to optimize: volume, and we want it largest, so a maximum.

Step 2:

V = \text{length} \times \text{width} \times \text{height}

Step 3: After cutting squares of side

x

from each corner:

Length of the base: $32 - 2x$ (you lose a square from each end)

Width of the base: $20 - 2x$

Height of the box: $x$ (once folded up)

V = (32 - 2x)(20 - 2x)(x)

Expand:

V = (640 - 104x + 4x^2)(x) = 4x^3 - 104x^2 + 640x

Result:

V(x) = 4x^3 - 104x^2 + 640x

, one variable.

Domain: since you’re cutting squares from corners,

0 < x < 10

(half of the shorter side).

Example 4: Minimizing Wire Length

Problem: Two vertical poles stand 40 feet apart. One is 20 feet tall, the other is 15 feet tall. A wire runs from the top of each pole to a single stake on the ground between them. Where do you put the stake to use the least wire?

What to optimize: total wire length, and we want the least, so a minimum.

Step 2:

W = W_1 + W_2

Step 3: Let

x

be the distance from the left pole to the stake, so

40 - x

is the distance from the stake to the right pole. Pythagorean theorem on each right triangle:

W_1 = \sqrt{20^2 + x^2} = \sqrt{400 + x^2}

W_2 = \sqrt{15^2 + (40-x)^2} = \sqrt{225 + (40-x)^2}

W = \sqrt{400 + x^2} + \sqrt{225 + (40-x)^2}

Result:

W(x) = \sqrt{400 + x^2} + \sqrt{225 + (40-x)^2}

, one variable.

Domain:

0 \leq x \leq 40

Key Strategy: Using Constraints

When you have two variables but need one:

Find the constraint equation, usually given in the problem

Solve the constraint for one variable in terms of the other

Substitute into the optimization equation

Pattern: optimize

A = xy

subject to

x + y = 50

From the constraint: $y = 50 - x$

Substitute: $A = x(50 - x) = 50x - x^2$

Now it’s in one variable

Common Optimization Setups

Scenario	What to Optimize	Common Formula
Area problems	Area	$A = lw$ or $A = \pi r^2$
Volume problems	Volume	$V = lwh$ or $V = \pi r^2 h$
Distance problems	Distance	$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Cost problems	Cost	$C = (\text{unit cost}) \times (\text{quantity})$
Revenue problems	Revenue	$R = (\text{price}) \times (\text{quantity})$
Profit problems	Profit	$P = R - C$

Setup Checklist

Before you take the derivative, make sure you’ve:

Identified what quantity to maximize or minimize

Drawn a picture if applicable (especially for geometry problems)

Labeled known quantities with numbers

Labeled unknown quantities with variables

Written the primary equation (the thing you’re optimizing)

Identified and used the constraint equation

Reduced everything to one variable only

Simplified the equation (optional but it helps)

Key Takeaways

Optimization means finding maximum or minimum values

Keywords to watch for: largest, smallest, most, least, greatest, shortest, maximum, minimum

Three setup steps: draw a picture, write the equation, convert to one variable

The primary equation is the quantity you want to optimize (area, volume, distance, cost, etc.)

The constraint equation is extra info that relates the variables

Use constraints to eliminate extra variables through substitution

You must have one variable before taking the derivative

The Pythagorean theorem is useful for distance and geometry problems

Domain matters, so think about physical constraints (lengths can’t be negative, etc.)

This lesson is setup only, solving comes next

Back Next