Optimization

Main idea

Most "what's the best way to..." questions turn into finding the highest or lowest point of a function. The hard part is writing that function in one variable; once you have it, it's the absolute extrema method you already know.

\text{maximize } {\htmlData{sym=0}{A}} = xy \text{ subject to } {\htmlData{sym=1}{2x + y = 500}} \;\Longrightarrow\; {\htmlData{sym=0}{A}}(x) = x(500 - 2x)

What the symbols mean

$A$
Objective
The quantity to make as large (or small) as possible.
$2x + y = 500$
Constraint
The fixed resource. Solve it for one variable and substitute.

Explanation

Optimization simply means finding the best possible choice: the biggest area, the cheapest design, the shortest route. All we're really doing is turning a word problem into a function and finding its highest or lowest point. The formula above has two pieces: the objective $A = xy$ (the thing you want as big or small as possible) and the constraint $2x + y = 500$ (the limit you're stuck with, like a fixed amount of fence). Use the constraint to get rid of one variable, and you're left with a one-variable function, $A(x) = x(500 - 2x)$ . From there it's a problem you've already solved.

Notice why the constraint is what makes the problem interesting. With unlimited fence, there's no "best" rectangle: you just keep making it bigger. The constraint forces a trade-off (a longer side means shorter ones elsewhere), and a trade-off is exactly what produces a peak somewhere in the middle.

The method:

Draw a picture and label the variables.
Write the objective: the quantity to optimize.
Write the constraint and use it to eliminate all but one variable.
Find the domain (lengths can't be negative, and so on).
Find the critical points and check them, with the endpoints or the second derivative test.
Answer the question that was asked, with units.

Most of the work is steps 1 to 3. Once you have a function of one variable, it's the closed interval method from the absolute extrema lesson again.

A common mistake. It's tempting to stop as soon as you find a critical point, because there's usually only one and it looks like the answer. Check it anyway: compare with the endpoints or use the second derivative, since a critical point can just as easily be the worst choice as the best. And reread the question, which might ask for the dimensions, the area, or the cost, not the $x$ you solved for.

See it

This is the volume of an open box made by cutting squares of side x from the corners of a 12 in sheet and folding up the sides, V(x) = x(12 − 2x)². Cut too little and the box is nearly flat; cut too much and there's hardly any base left. As you can see, the volume peaks in between, at x = 2 in, where the box holds 128 in³.

Worked examples

Example 1

A farmer has 500 ft of fence to enclose a rectangle along a river (no fence needed on the river side). What dimensions give the most area?

1
Call the two sides perpendicular to the river $x$ and the side along it $y$ . The fence covers three sides, so $2x + y = 500$ .

Example 2

An open box is made from a 12 in square sheet by cutting equal squares from the corners and folding up. What cut gives the largest volume?

1
After cutting squares of side $x$ , the base is $12 - 2x$ on each side and the height is $x$ , so $V(x) = x(12 - 2x)^2$ on $[0, 6]$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Two positive numbers add to 20. What is the largest possible product?

Problem 2+5 XP

A rectangle has perimeter 40. What is its largest possible area?

Problem 3+5 XP

An open box is made from an 18 in square sheet as in the example. What is the largest volume?

Problem 4+5 XP

Find the minimum value of $x + \dfrac1x$ for $x > 0$ .

Problem 5+5 XP

Which point on $y = \sqrt x$ is closest to $(3, 0)$ ? Give its x-coordinate.

Problem 6+5 XP

A farmer has $600$ m of fence to enclose a rectangular field along a straight river; the river side needs no fence. What's the largest area she can enclose?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A $2$ m statue stands on a $3$ m pedestal. A visitor whose eyes are $1.5$ m above the ground wants the statue to fill as large an angle of their view as possible. How far from the pedestal should they stand?

Where it's used

Packaging that uses the least material, routes that save the most fuel, prices that bring in the most profit: optimization is calculus's most direct payoff.

Maximizing ProfitCalc I · Economics & Finance Best Choice on a BudgetCalc III · Economics & Finance Light Takes the Fastest PathCalc I · Physics

The Mean Value Theorem More Optimization Problems