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

Optimization: Minimizing Cost and Materials

B2.4About 70 minutes

Learning goals

I can minimize surface area for a container of fixed volume.
I can build a cost function when different parts have different prices.
I can optimize simple rational functions of the form ax² + b/x and av + b/v.
I can minimize a cost that involves a distance found with the Pythagorean theorem.

Words to know

surface area
The total area of all the faces of a solid; for a closed cylinder, 2πr² + 2πrh.

unit cost
The cost per unit of something, such as dollars per square metre or per kilometre.

cost function
A function that gives the total cost in terms of a chosen variable.

simple rational function
A function such as 2πr² + 710/r that includes a power with a negative exponent.

Less metal, same pop

A standard Canadian pop can holds 355 mL. Every can needs aluminum for its top, bottom and side, and a beverage company makes hundreds of millions of cans. Shaving even a little metal off each one saves real money and energy. So what shape of cylinder holds 355 mL with the least surface area?

In the last lesson you maximized areas and volumes. This lesson minimizes: material, cost, fuel. Many of these problems lead to simple rational functions like S(r) = 2πr² + 710/r, exactly the kind of function B2.4 asks you to optimize.

Worked example

Example 1: The most efficient can

Find the radius and height of a closed cylindrical can that holds 355 cm³ (355 mL) and uses the least material.

Surface area (top + bottom + side): S = 2πr² + 2πrh.
Constraint: πr²h = 355, so h = 355/(πr²).
Substitute: S(r) = 2πr² + 2πr × 355/(πr²) = 2πr² + 710/r, for r > 0.
S′(r) = 4πr − 710/r² = 0 gives 4πr³ = 710, so r³ = 710/(4π) ≈ 56.50 and r ≈ 3.84 cm.
Only one critical number on r > 0. S′(1) < 0 and S′(5) > 0, so it is a local and therefore absolute minimum.
h = 355/(π × 3.837²) ≈ 7.67 cm. Notice h ≈ 2r: the height equals the diameter.

r ≈ 3.84 cm and h ≈ 7.67 cm, using about 277 cm² of aluminum. Real pop cans are taller and narrower than this, because other things matter too: how the can fits in a hand, how the thicker top is made, and how cans pack and stack.

Left: a can with radius r and height h. Right: the surface area S(r) = 2πr² + 710/r of a 355 mL can. Very thin cans need a huge side; very wide cans need huge ends. The minimum is in between.

Worked example

Example 2: Different prices for different faces

A closed storage bin with a square base must hold 2 m³. The base and the lid cost $10/m², and the sides cost $6/m². Find the dimensions that minimize the cost.

Base x by x, height h. Constraint: x²h = 2, so h = 2/x².
Cost: base and lid 10(2x²), four sides 6(4xh). C = 20x² + 24x(2/x²) = 20x² + 48/x, for x > 0.
C′(x) = 40x − 48/x² = 0 gives x³ = 1.2, so x ≈ 1.063 m.
One critical number; C′ changes from negative to positive there, so it is the absolute minimum.
h = 2/x² ≈ 2/1.129 ≈ 1.771 m. C ≈ 20(1.129) + 48/1.063 ≈ $67.75.

Base about 1.06 m by 1.06 m, height about 1.77 m, for a minimum cost of about $67.75. Because the base and lid are pricier, the best bin is taller than a cube.

Worked example

Example 3: Laying a power cable

A power station is on one bank of a straight river 1 km wide. A factory is on the other bank, 4 km downstream. A cable will run x km along the station's bank (at $3000/km) and then underwater straight to the factory (at $5000/km). Find x to minimize the cost.

Underwater length: the cable crosses 1 km and covers the remaining (4 − x) km downstream, so it is √(1 + (4 − x)²) km.
C(x) = 3000x + 5000(1 + (4 − x)²)^(1/2), for 0 ≤ x ≤ 4.
Chain rule: C′(x) = 3000 + 5000 × (1/2)(1 + (4 − x)²)^(−1/2) × 2(4 − x)(−1) = 3000 − 5000(4 − x)/√(1 + (4 − x)²).
Set C′ = 0 and let u = 4 − x: u/√(1 + u²) = 0.6. Square: u² = 0.36 + 0.36u², so 0.64u² = 0.36 and u = 0.75. Then x = 3.25.
Compare: C(0) = 5000√17 ≈ $20 616; C(3.25) = 9750 + 5000(1.25) = $16 000; C(4) = 12 000 + 5000 = $17 000.

Run the cable 3.25 km along the bank, then 1.25 km underwater, for a minimum cost of $16 000.

The cable route of Example 3: x km along the near bank, then underwater straight to the factory. The best route crosses at x = 3.25 km. (Not exactly to scale.)

Worked example

Example 4: The cheapest driving speed

A delivery truck makes a 400 km trip. The driver earns $36 per hour, and fuel costs 0.01v² dollars per hour at a speed of v km/h. The speed must be between 40 and 100 km/h. Find the speed that minimizes the cost of the trip.

Time = 400/v hours. Cost per hour = 36 + 0.01v².
C(v) = (400/v)(36 + 0.01v²) = 14 400/v + 4v, for 40 ≤ v ≤ 100.
C′(v) = −14 400/v² + 4 = 0 gives v² = 3600, so v = 60.
Compare: C(40) = 360 + 160 = $520; C(60) = 240 + 240 = $480; C(100) = 144 + 400 = $544.

Drive at 60 km/h for a minimum trip cost of $480. Faster saves wages but burns more fuel; slower does the opposite.

A pattern worth noticing

In Examples 1, 2 and 4 the function had the form ax² + b/x or av + b/v. At the minimum, the two parts are balanced in a simple way. For instance, in Example 4 the wage cost and fuel cost are equal ($240 each) at the best speed. This is common in cost problems, and a nice check on your answer.

Just for fun

If pop cans were the "most efficient" shape (height = diameter), they would look like little drums. Calculus says it would save metal; your hand says it would be awkward to hold. Engineers listen to both.

Hands-on activity

Pantry can survey

You need: 4 to 6 different cans from your kitchen (soup, tuna, tomatoes, pop), a ruler, and a calculator.

For each can, measure the diameter d and height h in centimetres, and read the volume on the label (in mL; 1 mL = 1 cm³).
Calculate the ratio h ÷ d for each can.
For one can, use its volume V to find the most efficient radius r = ∛(V/(2π)), and compare with the real radius.
Discuss: which cans are close to h = d? Why might the others be different?

What to look for: soup cans are usually somewhat taller than they are wide, tuna cans are much flatter (h ÷ d well below 1) and pop cans are much taller (h ÷ d near 2). Reasons include how food is packed and cooked in the can, labels, shelf space, holding the can, and the different thickness of tops and sides.

Your turn

A closed cylinder must hold 1000 cm³. Find the radius and height that minimize surface area.

An open-top cylindrical cup must hold 500 cm³. Find the dimensions that use the least material.

An open-top box with a square base must hold 4 m³. The base costs $8/m² and the sides $4/m². Find the dimensions of least cost.

A trip's cost is C(v) = 22 500/v + 4v dollars. Find the cheapest speed.

A 5000 m² rectangular lot has one side on a road. Fencing on the road side costs $12/m and on the other three sides $8/m. Find the dimensions of least cost.

Find the minimum of f(x) = 2x + 50/x for x > 0.

Redo the cable problem with the factory 3 km downstream, land cable at $4000/km and underwater at $5000/km (river still 1 km wide).

More practice: warm-up

Write the volume of a cylinder.

Write the total surface area of a closed cylinder.

If πr²h = 100, write h in terms of r.

Differentiate 200/r.

Material costs $12/m². What do 3 m² cost?

More practice: on your own

A closed box with a square base must hold 1000 cm³. The top and bottom cost twice as much per cm² as the sides. Find the cheapest dimensions.

A 355 mL can has its top and bottom cost twice as much per cm² as its side. Find the cheapest radius and height.

A tour boat's fuel costs 0.2v² dollars per hour at v km/h, and the crew costs $80 per hour. Find the cheapest speed for a 100 km trip.

A 600 m² storage yard has a front fence costing $20/m and the other three sides $10/m. Find the cheapest dimensions.

A marketing campaign lasting t weeks has total cost C(t) = 20t + 500e^(−0.1t) thousand dollars, where the second term is the cost of unsold stock. Find the campaign length that minimizes cost.

Challenge

Prove that a closed cylinder of any fixed volume V has the least surface area when its height equals its diameter.

A truck makes a 600 km trip at v km/h (60 ≤ v ≤ 100). Wages are $27/h and fuel costs v²/300 dollars per hour. Find the cheapest speed and cost.

A closed box with a square base holds volume V. The top and bottom cost k times as much per unit area as the sides. Show that the cheapest box has height h = kx, where x is the side of the base.

Check your understanding

Quiz

A closed cylinder holds 355 cm³. Which function gives its surface area in terms of r?

For a closed cylinder of fixed volume made of uniform material, the minimum surface area occurs when…

C(v) = 14 400/v + 4v. Where is C′(v) = 0 for v > 0?

In Example 3, why must C(0) and C(4) be checked?

C(x) = 20x² + 48/x for x > 0. What is C′(x)?

When the base and lid of a box cost more per m² than its sides, the cheapest box with a square base is…

For parents and teachers

Big idea (B2.4): minimizing material or cost works just like maximizing area: write the cost as a function of one variable using the constraint, then find and test the critical numbers. Many of these functions are simple rational functions.

How to help: do the pantry can survey together, and talk about why real cans are not all the "best" shape. This builds the habit of checking whether a mathematical answer is realistic.

Common difficulties:

Forgetting the top and bottom of a closed container (or including a top that is open).
Multiplying each area by the wrong unit cost.
Errors differentiating b/x (it is −b/x²).
Not checking the endpoints when the domain is closed (Examples 3 and 4).