Answer the original question
The price of a tour
A tour company charges 50 dollars per person, and 80 people come. For each 5-dollar increase in the price, 4 fewer people come. Let \(x\) be the number of 5-dollar increases. The company asks what price to charge.
Drag \(x\). The readouts show the price, the people and the revenue, and the graph shows the revenue \(R(x)\).
The graph is highest at \(x = 5\). But 5 counts price increases, so it is not a price.
Move \(x\) to 5. The readouts show the answer: a price of 75 dollars, 60 people and a revenue of 4500 dollars.
The input is a step, not the answer
The input \(x\) sets every quantity the question asks about: the price, the number of people and the revenue.
Drag the input. The revenue, the height of the graph, changes with it, and so do the price and the number of people. The best input is only a step: the answer is these quantities at that input, each with its unit.
You write the answer with the words of the question.
Notation
Reread the question and list every quantity it asks for. The best input \(c\) is on the axis, not in the list.
Substitute \(c\) into the relations of the model. Each relation gives one asked quantity.
\[\begin{aligned} \cy{\text{price}} &= 50 + 5c \\ \cy{\text{people}} &= 80 - 4c \end{aligned}\]Write each answer with its unit. Give an exact value first, and add a decimal only if the question needs one.
Test the answer: \(c\) must lie in the domain, and each quantity must make sense in the situation.
\[0 \le c \le 20\]State the answer in a sentence that uses the words of the question.
When it fails
The best input \(c = 5\) is not the answer. The red sentence gives a price of 5 dollars, but the price at \(c\) is \(50 + 5 \cdot 5 = 75\) dollars.
An answer that gives only the revenue, 4500 dollars, also fails, because the question asks for the price.
Example 1: A window
A window is a rectangle topped by a semicircle of radius \(r\) m. The distance around the outside is 12 m. The area is \(A(r) = 12r - \left(2 + \frac{\pi}{2}\right)r^2\), and it is greatest at \(r = \frac{12}{4 + \pi}\). Give the width and the height of the rectangle, and the area.
The question asks for the width, the height and the area, not for \(r\). The width is twice the radius.
\[\text{width} = 2r = \frac{24}{4 + \pi}\ \text{m}\]The distance around gives the height, because the two sides, the bottom and the arc add up to 12 m.
\[\begin{aligned} 2h + 2r + \pi r &= 12 \\ h &= 6 - r - \frac{\pi r}{2} \\ &= 6 - \frac{6(2 + \pi)}{4 + \pi} \\ &= \frac{12}{4 + \pi}\ \text{m} \end{aligned}\]The area is the value of \(A(r)\) at the best \(r\).
\[\begin{aligned} A &= 12r - \left(2 + \frac{\pi}{2}\right)r^2 \\ &= \frac{144}{4 + \pi} - \frac{72}{4 + \pi} \\ &= \frac{72}{4 + \pi}\ \text{m}^2 \end{aligned}\]The rectangle is \(\frac{24}{4 + \pi}\) m wide and \(\frac{12}{4 + \pi}\) m tall, and the window has area \(\frac{72}{4 + \pi}\) m². As a check, these are about 3.36 m, 1.68 m and 10.08 m².
Example 2: A gutter
A strip of metal 24 cm wide is folded into a gutter. A strip of width \(x\) cm turns up along each edge. The area of the cross-section is \(A(x) = x(24 - 2x)\), and it is greatest at \(x = 6\). Give the depth and the bottom width of the gutter, and the area.
The strips that turn up are the sides of the gutter, so \(x\) is the depth.
A tempting wrong move: The gutter is 6 cm wide.
\[\text{width} = x = 6\ \text{cm}\]The number 6 is the depth, the orange side. The bottom is twice as long as each side, so the red 6 cm does not fit it.
The bottom is the part of the strip that is not turned up, the chalk piece of the bar below.
\[\begin{aligned} \text{bottom width} &= 24 - 2x \\ &= 12\ \text{cm} \end{aligned}\]The gutter is 6 cm deep and 12 cm wide at the bottom. Its cross-section has area 72 cm².
\[A(6) = 6 \cdot 12 = 72\ \text{cm}^2\]
Practice
A rectangle is inscribed in a circle of radius 5 m. Its half-width is \(x\) m, so its area is \(A(x) = 4x\sqrt{25 - x^2}\), for \(0 \le x \le 5\). The area is greatest at \(x = \frac{5\sqrt2}{2}\). Give the dimensions and the area of the largest rectangle.
The width is \(2x\) and the height is \(2\sqrt{25 - x^2}\). Substitute the best \(x\) into both.
Solution
The width is twice the half-width.
\[2x = 5\sqrt2\ \text{m}\]At the best \(x\), \(x^2 = \frac{25}{2}\), so \(25 - x^2 = \frac{25}{2}\) as well. The height is then the same as the width.
\[2\sqrt{25 - x^2} = 2\sqrt{\tfrac{25}{2}} = 5\sqrt2\ \text{m}\]The width times the height gives the area. The largest rectangle is a square with side \(5\sqrt2\) m and area 50 m².
\[5\sqrt2 \cdot 5\sqrt2 = 50\ \text{m}^2\]
Questions
Questions 1 and 2 use an orchard. A farmer plants \(n\) extra trees per acre on top of 40, and each tree then gives \(600 - 10n\) apples. The graph shows the total yield \(T(n)\) in apples per acre.
The graph is highest at \(n = 10\). How many trees per acre should the farmer plant?
- 50 trees per acre
- 10 trees per acre
- 500 trees per acre
- 25,000 trees per acre
Answer
50 trees per acre
At the top of the graph, how many apples does each tree give?
- 500 apples
- 600 apples
- 10 apples
- 25,000 apples
Answer
500 apples
A rectangle has area 72 cm² and width \(x\) cm. Its perimeter is \(P(x) = 2x + \dfrac{144}{x}\), and it is least at \(x = 6\sqrt2\). What is the least perimeter?
- \(24\sqrt2\) cm
- \(6\sqrt2\) cm
- \(12\sqrt2\) cm
- 72 cm
Answer
\(24\sqrt2\) cm
Summary
You substitute the best input into the model and write every quantity the question asks for, with its unit.
This works because the best input is only a step: the readouts at \(x = 5\) show the answer, the price, the people and the revenue.
Use it as the last step of every optimization problem.
This is one of four skills in topic 5.11 of Which & Why: AP Calculus.