B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and Optimization

Optimization in Geometry: Areas and Volumes

B2.4About 70 minutes

Learning goals

I can follow a step-by-step strategy to solve optimization problems.
I can write an area or volume as a function of one variable using a constraint.
I can state a realistic domain and use it to choose the correct critical number.
I can maximize areas and volumes and minimize lengths in geometric problems.

Words to know

optimization
Finding the largest or smallest possible value of a quantity, and the input that produces it.

objective function
The quantity to be maximized or minimized, written as a function.

constraint
A condition the variables must satisfy, such as a fixed perimeter or a fixed volume.

feasible domain
The set of input values that make sense in the problem.

How much can you fence in?

A farmer near the Grand River has 600 m of fencing and wants a rectangular pasture beside the river. The river side needs no fence. What shape gives the cows the most grass?

Try a few: 100 m deep makes the long side 400 m, for an area of 40 000 m². 200 m deep leaves 200 m, for 40 000 m² again. Somewhere between is a better choice. Guessing works, slowly. Calculus finds the exact best answer in a few lines.

A strategy for optimization problems

Understand. Read the problem twice. What quantity is to be maximized or minimized? What is fixed?
Draw and label a diagram. Use letters for quantities that can change.
Write the quantity to optimize as a formula (area, volume, cost...).
Use the constraint (the fixed amount) to write it as a function of one variable.
State the domain: which values of the variable make physical sense?
Differentiate, find critical numbers, and use the closed-interval method or a derivative test from the last lesson.
Answer the question that was asked, in a sentence with units. (It may ask for the dimensions, not the area.)

Worked example

Example 1: The riverside pasture

A farmer has 600 m of fencing for a rectangular pasture along a straight river. No fence is needed along the river. Find the dimensions that give the largest area.

Let x be the length of each side perpendicular to the river. The side parallel to the river is 600 − 2x.
Area: A(x) = x(600 − 2x) = 600x − 2x².
Domain: both lengths must be non-negative, so 0 ≤ x ≤ 300.
A′(x) = 600 − 4x = 0 at x = 150.
Closed-interval check: A(0) = 0, A(150) = 150(300) = 45 000, A(300) = 0.

The pasture should be 150 m deep and 300 m along the river, for a maximum area of 45 000 m². (The side along the river is twice the depth.)

The pasture of Example 1. Only three sides need fencing: two sides of length x and one side of length 600 − 2x.

Worked example

Example 2: An open-top box

A box with an open top is made from a 30 cm by 20 cm sheet of cardboard by cutting an x-by-x square from each corner and folding up the sides. Find the value of x that gives the largest volume.

After cutting and folding, the box is x cm high, its base is (30 − 2x) cm by (20 − 2x) cm.
V(x) = x(30 − 2x)(20 − 2x) = 4x³ − 100x² + 600x.
Domain: 20 − 2x ≥ 0, so 0 ≤ x ≤ 10.
V′(x) = 12x² − 200x + 600 = 0, or 3x² − 50x + 150 = 0. Quadratic formula: x = [50 ± √(2500 − 1800)] ÷ 6 = (50 ± √700) ÷ 6.
x ≈ (50 − 26.458) ÷ 6 ≈ 3.92 or x ≈ 12.74. Only 3.92 is in the domain.
V(0) = 0, V(10) = 0, V(3.924) ≈ 3.924 × 22.153 × 12.153 ≈ 1056.

Cut squares of about 3.92 cm. The box then holds about 1056 cm³ (just over a litre).

Cutting x-by-x squares from the corners of a 30 cm by 20 cm sheet and folding up the sides gives a box of height x with a (30 − 2x) by (20 − 2x) base. (Not to scale.)

The volume V(x) for 0 ≤ x ≤ 10. Small cuts make a flat tray; big cuts make a tall, skinny box. The best box is in between, where V′(x) = 0.

Worked example

Example 3: A rectangle under a parabola

A rectangle has its base on the x-axis and its two upper corners on the parabola y = 12 − x². Find the largest possible area.

By symmetry, let the upper right corner be (x, 12 − x²), with x > 0. The width is 2x and the height is 12 − x².
A(x) = 2x(12 − x²) = 24x − 2x³, with domain 0 ≤ x ≤ √12.
A′(x) = 24 − 6x² = 0 at x = 2 (x > 0).
A(0) = 0, A(√12) = 0, A(2) = 4 × 8 = 32.

The largest rectangle is 4 wide and 8 tall, with area 32 square units.

Worked example

Example 4: Least fencing for a given area

A rectangular garden of 200 m² is to be built against the wall of a barn, so only three sides need fencing. What dimensions use the least fencing?

Let x be the length of each side perpendicular to the wall, and y the side parallel to it. Constraint: xy = 200, so y = 200/x.
Fencing: L(x) = 2x + y = 2x + 200/x = 2x + 200x⁻¹, for x > 0.
L′(x) = 2 − 200/x² = 0 gives x² = 100, so x = 10.
Only one critical number on x > 0; L′(5) = 2 − 8 < 0 and L′(20) = 2 − 0.5 > 0, so it is the absolute minimum.
Then y = 200/10 = 20 and L = 20 + 20 = 40.

Make the garden 10 m out from the wall and 20 m along it, using 40 m of fencing.

Just for fun

Calculus says the best riverside pasture is twice as long as it is deep. The cows have not read the proof, but they agree with the result.

Hands-on activity

Build the best paper box

You need: 6 sheets of letter paper (21.6 cm by 27.9 cm), a ruler, a pencil, scissors (use them carefully), tape, and a calculator.

On the sheets, mark corner squares of side x = 1, 2, 3, 4, 5 and 6 cm (one value per sheet). Cut out the four corner squares and fold up the sides. Tape the corners.
For each box, measure its length, width and height, and calculate its volume.
Which cut gave the biggest box?
Now write V(x) = x(27.9 − 2x)(21.6 − 2x), find V′(x), and solve V′(x) = 0 with the quadratic formula.

What to look for: the volumes should be roughly 508, 841, 1025, 1083, 1038 and 916 cm³ for x = 1 to 6, so the best whole-centimetre cut is 4 cm. Calculus gives x ≈ 4.03 cm and V ≈ 1083 cm³. Your measured boxes will be a little off because of the folds and the tape.

Your turn

Two positive numbers add to 20. Find them so that their product is as large as possible.

A rectangle has a perimeter of 80 m. Find the largest possible area.

With 400 m of fence, a rectangular field is fenced on three sides along a straight stream. Find the maximum area.

An open box is made from a 12 cm by 12 cm square by cutting x-by-x squares from the corners. Find the largest volume.

A rectangle has its base on the x-axis and upper corners on y = 27 − x². Find its largest area.

A 450 m² rectangular pen is built against a barn wall. Find the least amount of fencing for the other three sides.

A farmer has 1200 m of fence to enclose a rectangle and divide it into two pens with a fence parallel to one side. Find the largest total area.

More practice: warm-up

What is the first step in solving an optimization problem?

A rectangle has perimeter 100 and width x. Write its area as a function of x.

State the domain for A(x) = x(50 − x) in this context.

Find A′(x) for A(x) = x(50 − x).

An open box is made from a 10 by 10 sheet by cutting x-by-x corners. Write V(x).

More practice: on your own

An open-top box with a square base must hold 32 000 cm³. Find the dimensions that use the least material.

A rectangle is inscribed in a semicircle of radius 10, with its base on the diameter. Find the largest area.

A field along a river is fenced on three sides with 2400 m of fence and divided in two by a fence perpendicular to the river. Find the maximum area.

Find the point on the line y = 10 − 2x closest to the origin.

A poster must have 600 cm² of printed area with 3 cm margins at top and bottom and 2 cm margins at each side. Find the poster dimensions that use the least paper.

Challenge

A rectangle has its base on the x-axis and its upper corners on y = e^(−x²). Find the largest area.

A Norman window is a rectangle topped by a semicircle. Its total perimeter is 8 m. Find the radius of the semicircle that lets in the most light (largest area).

Prove that of all rectangles with perimeter P, the square has the largest area.

Check your understanding

Quiz

In an optimization problem, what is the constraint used for?

A rectangle has perimeter 60 cm. What is its largest possible area?

With 200 m of fence for three sides of a rectangle along a river, which depth (perpendicular to the river) gives the largest area?

An open box is made from a 10 cm by 16 cm sheet by cutting x-by-x squares from the corners. What is the domain of V(x)?

Why do we check the endpoints of the domain in a box problem like Example 2?

A 200 m² garden against a wall needs fence on three sides. L(x) = 2x + 200/x. Where is L′(x) = 0?

For parents and teachers

Big idea (B2.4): an optimization problem becomes a calculus problem once you write the objective as a function of one variable on a sensible domain. The derivative finds the candidates; the closed-interval method or a derivative test picks the winner.

How to help: ask your teen to show you the diagram and to say in words what is fixed and what can change. Building the paper box in the activity together is a good way to see the trade-off.

Common difficulties:

Not using the constraint, leaving two variables.
Forgetting the domain, or keeping a critical number outside it.
Answering with the maximum area when the question asks for the dimensions (or the other way around).
Algebra slips when expanding products like x(30 − 2x)(20 − 2x).