Identify the quantity to optimize
A wire cut in two
A wire 40 cm long is cut into two pieces, and each piece is bent into a square. The cut \(x\) is the length of the orange piece, and the two squares are shaded teal.
Drag the cut. The two pieces always add up to 40 cm, so the wire length does not change, but the total area of the squares does.
At a cut of 20 cm, each square has a side of 5 cm, so the total area is 50 cm².
Suppose the question asks for the cut with the smallest total area. The area is then the quantity to make smallest, and the 40 cm is an amount that every cut must respect.
Any wire
Take a wire of any length \(L\). Drag the cut. The total area changes, and the length \(L\) does not.
The quantity that the question asks you to make largest or smallest is the objective, here the total area. The amount that stays the same for every cut is the constraint, here the length \(L\).
Notation
The total area is the objective, because the question asks for the smallest area. Words such as largest, least and cheapest point to the objective.
\[\cb{A} = \text{total area}\]The wire length is a constraint, because the problem fixes it at 40 cm for every cut. Words such as has, must and a total of point to a constraint.
\[\cq{\text{wire length}} = 40\]To test a quantity, change the shape. If it changes, it can be the objective, and if it stays the same, it is a constraint. Name both in words before you write a formula.
When it fails
Choose the wire length as the quantity to make smallest. Its graph is a flat line at 40.
Every cut gives 40 cm, so no cut is best. The objective must be a quantity that changes when the shape changes.
Example 1: Two numbers
Two numbers add to 20. The product of the numbers is to be as large as possible. Name the objective and the constraint.
Drag the point. The numbers 5 and 15 have product 75, and 10 and 10 have product 100, so the product changes.
\[5 \cdot 15 = 75, \qquad 10 \cdot 10 = 100\]The sum is 20 for every choice of the two numbers, so it is the constraint.
\[x + y = \cq{20}\]The question asks for the largest product, so the product is the objective.
\[\cb{P} = xy\]
Example 2: The cheapest cable
A plant is on one bank of a river 3 km wide. A factory is on the far bank, 4 km down the river from the point straight across. A cable crosses the river and then follows the bank. It costs $5,000 per km under water and $3,000 per km on land. Find the cheapest cable.
The cable lands at a point \(x\) km down the bank from the point straight across. A different \(x\) gives a different cable, so the cost changes.
A tempting wrong move: Use the shortest cable. It goes straight to the factory, with \(x = 4\).
\[\text{length} = \sqrt{3^2 + 4^2} = 5 \text{ km}\]The question asks for the cheapest cable, not the shortest. This cable is under water for 5 km and costs $25,000. The graph is higher at \(x = 4\) than at its lowest point.
The question asks for the cheapest cable, so the objective is the cost \(C(x)\). The part under water is the hypotenuse of a right triangle with legs 3 and \(x\), and the part on land is \(4 - x\) long.
\[\begin{aligned} \cb{C(x)} &= 5000\sqrt{9 + x^2} \\ &\quad + 3000(4 - x) \end{aligned}\]The 3 km width and the 4 km along the bank are the same for every cable, so they are the constraints.
Practice
A shop sells \(x\) posters a day. When it sells \(x\) posters, the price of each poster is \(p = 20 - \tfrac{x}{2}\) dollars. The shop wants the largest daily income. Name the objective and the constraint.
Ask which quantity the shop wants to make largest. Then ask which fact holds for every number of posters.
Solution
The shop wants the largest income, so the income \(R\), the number of posters times the price, is the objective.
\[\cb{R} = x \cdot p\]The number sold and the price cannot be chosen separately, because the rule \(p = 20 - \tfrac{x}{2}\) links them. The rule holds for every \(x\), so it is the constraint.
\[\begin{aligned} x = 10&: \ p = 15,\ R = 150 \\ x = 20&: \ p = 10,\ R = 200 \end{aligned}\]
Questions
A farmer has 100 m of fence for a pen with a barn wall on one side. The two sides next to the wall are \(x\) long.
The farmer wants the largest pen. Which quantity is the objective?
- The area of the pen
- The length of fence
- The side \(x\)
- The side along the wall
Answer
The area of the pen
Which quantity is the same for every position of the point?
- The total length of fence, 100 m
- The area of the pen
- The side \(x\)
- The side along the wall
Answer
The total length of fence, 100 m
A rectangle has a perimeter of 60 cm. You want its diagonal as short as possible. Which statement is correct?
- The diagonal is the objective, and the perimeter is a constraint
- The perimeter is the objective, and the diagonal is a constraint
- The area is the objective
- There is no objective, because the perimeter is fixed
Answer
The diagonal is the objective, and the perimeter is a constraint
Summary
You name the quantity to make largest or smallest as the objective, and the fixed amounts as constraints.
This works because the objective changes when the shape changes, and a constraint is the same for every shape that fits the problem.
You use it first in any optimization problem, before you write a formula.
This is one of four skills in topic 5.10 of Which & Why: AP Calculus.