4.7 Applied optimization problems (Page 4/8)

Calculus volume 1 Page 4 / 8

In business, companies are interested in maximizing revenue . In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let’s use these data to determine the price the company should charge to maximize the amount of money it brings in.

Maximizing revenue

Owners of a car rental company have determined that if they charge customers $p$ dollars per day to rent a car, where $50 \leq p \leq 200,$ the number of cars $n$ they rent per day can be modeled by the linear function $n (p) = 1000 - 5 p .$ If they charge $$ 50$ per day or less, they will rent all their cars. If they charge $$ 200$ per day or more, they will not rent any cars. Assuming the owners plan to charge customers between $50 per day and $$ 200$ per day to rent a car, how much should they charge to maximize their revenue?

Step 1: Let $p$ be the price charged per car per day and let $n$ be the number of cars rented per day. Let $R$ be the revenue per day.

Step 2: The problem is to maximize $R .$

Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day—that is, $R = n \times p .$

Step 4: Since the number of cars rented per day is modeled by the linear function $n (p) = 1000 - 5 p,$ the revenue $R$ can be represented by the function

R (p) = n \times p = (1000 - 5 p) p = −5 p^{2} + 1000 p .

Step 5: Since the owners plan to charge between $$ 50$ per car per day and $$ 200$ per car per day, the problem is to find the maximum revenue $R (p)$ for $p$ in the closed interval $[50, 200] .$

Step 6: Since $R$ is a continuous function over the closed, bounded interval $[50, 200],$ it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is $R^{'} (p) = −10 p + 1000 .$ Therefore, the critical point is $p = 100$ When $p = 100,$ $R (100) = $ 50,000 .$ When $p = 50,$ $R (p) = $ 37,500 .$ When $p = 200,$ $R (p) = $ 0 .$ Therefore, the absolute maximum occurs at $p = $ 100 .$ The car rental company should charge $$ 100$ per day per car to maximize revenue as shown in the following figure.

The function R(p) is graphed. At its maximum there is an intersection of two dashed lines and text that reads “Maximum revenue is $50,000 per day when the price charged per car is $100 per day.” — To maximize revenue, a car rental company has to balance the price of a rental against the number of cars people will rent at that price.

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A car rental company charges its customers $p$ dollars per day, where $60 \leq p \leq 150 .$ It has found that the number of cars rented per day can be modeled by the linear function $n (p) = 750 - 5 p .$ How much should the company charge each customer to maximize revenue?

The company should charge $$ 75$ per car per day.

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Maximizing the area of an inscribed rectangle

A rectangle is to be inscribed in the ellipse

\frac{x^{2}}{4} + y^{2} = 1 .

What should the dimensions of the rectangle be to maximize its area? What is the maximum area?

Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let $L$ be the length of the rectangle and $W$ be its width. Let $A$ be the area of the rectangle.

The ellipse x2/4 + y2 = 1 is drawn with its x intercepts being ±2 and its y intercepts being ±1. There is a rectangle inscribed in the ellipse with length L (in the x-direction) and width W. — We want to maximize the area of a rectangle inscribed in an ellipse.

Step 2: The problem is to maximize $A .$

Step 3: The area of the rectangle is $A = L W .$

Step 4: Let $(x, y)$ be the corner of the rectangle that lies in the first quadrant, as shown in [link] . We can write length $L = 2 x$ and width $W = 2 y .$ Since $\frac{x^{2}}{4 + y^{2} = 1}$ and $y > 0,$ we have $y = \sqrt{\frac{1 - x^{2}}{4}} .$ Therefore, the area is

A = L W = (2 x) (2 y) = 4 x \sqrt{\frac{1 - x^{2}}{4}} = 2 x \sqrt{4 - x^{2}} .

Step 5: From [link] , we see that to inscribe a rectangle in the ellipse, the $x$ -coordinate of the corner in the first quadrant must satisfy $0 < x < 2 .$ Therefore, the problem reduces to looking for the maximum value of $A (x)$ over the open interval $(0, 2) .$ Since $A (x)$ will have an absolute maximum (and absolute minimum) over the closed interval $[0, 2],$ we consider $A (x) = 2 x \sqrt{4 - x^{2}}$ over the interval $[0, 2] .$ If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.

Step 6: As mentioned earlier, $A (x)$ is a continuous function over the closed, bounded interval $[0, 2] .$ Therefore, it has an absolute maximum (and absolute minimum). At the endpoints $x = 0$ and $x = 2,$ $A (x) = 0 .$ For $0 < x < 2,$ $A (x) > 0 .$ Therefore, the maximum must occur at a critical point. Taking the derivative of $A (x),$ we obtain

\begin{matrix} A' (x) & = 2 \sqrt{4 - x^{2}} + 2 x \cdot \frac{1}{2 \sqrt{4 - x^{2}}} (−2 x) \\ = 2 \sqrt{4 - x^{2}} - \frac{2 x^{2}}{\sqrt{4 - x^{2}}} \\ = \frac{8 - 4 x^{2}}{\sqrt{4 - x^{2}}} . \end{matrix}

To find critical points, we need to find where $A' (x) = 0 .$ We can see that if $x$ is a solution of

\frac{8 - 4 x^{2}}{\sqrt{4 - x^{2}}} = 0,

then $x$ must satisfy

8 - 4 x^{2} = 0 .

Therefore, $x^{2} = 2 .$ Thus, $x = ± \sqrt{2}$ are the possible solutions of [link] . Since we are considering $x$ over the interval $[0, 2],$ $x = \sqrt{2}$ is a possibility for a critical point, but $x = − \sqrt{2}$ is not. Therefore, we check whether $\sqrt{2}$ is a solution of [link] . Since $x = \sqrt{2}$ is a solution of [link] , we conclude that $\sqrt{2}$ is the only critical point of $A (x)$ in the interval $[0, 2] .$ Therefore, $A (x)$ must have an absolute maximum at the critical point $x = \sqrt{2} .$ To determine the dimensions of the rectangle, we need to find the length $L$ and the width $W .$ If $x = \sqrt{2}$ then

y = \sqrt{1 - \frac{{(\sqrt{2})}^{2}}{4}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}} .

Therefore, the dimensions of the rectangle are $L = 2 x = 2 \sqrt{2}$ and $W = 2 y = \frac{2}{\sqrt{2}} = \sqrt{2} .$ The area of this rectangle is $A = L W = (2 \sqrt{2}) (\sqrt{2}) = 4 .$

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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