Chapter 3.3, Problem 80AYU

Business The daily revenue $R$ achieved by selling $x$ boxes of candy is figured to be $R\left(x\right)=9.5x-0.04x^2$. The daily cost $C$ of selling $x$ boxes of candy is $C\left(x\right)=1.25x+250$.

(a) How many boxes of candy must the firm sell to maximize revenue? What is the maximum revenue?

(b) Profit is given as $P(x)=R(x)–C\left(x\right)$. What is the profit function?

(c) How many boxes of candy must the firm sell to maximize profit? What is the maximum profit?

(d) Provide a reasonable explanation as to why the answers found in parts $(a)\text{ and }(c)$ differ. Explain why a quadratic function is a reasonable model for revenue.

To determine

To calculate:

The maximum revenue and the number of boxes of candy to be sold to maximise the revenue.

Profit function

The maximum profit and the number of boxes of candy the firm has to sell in order to maximise the profit.

The difference between parts (a) and (c) and why is quadratic function a more reasonable model for revenue.

Answer to Problem 80AYU

The maximum revenue is $\$564.06$ and is obtained by selling 119 candy boxes.

$P(x)=-0.04x^2+8.25x-250$

The maximum profit is $\$1755.39$ and is obtained by selling 103 candy boxes.

The answer is given below.

Explanation of Solution

Given:

The monthly revenue achieved by selling $x$ wrist watches is

$R(x)=9.5x-0.04x^2$

The monthly cost of selling $x$ wrist watches is

$C(x)=1.25x+250$

The profit function is $P(x)=R(x)-C(x)$

Formula used:

Consider a quadratic function of the form $f(x)=a{x}^2+bx+c,a\ne 0$.

Then the graph of the above function is a parabola with vertex $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$.

This vertex is the highest point if $a<0$ and the lowest point if $a>0$.

Therefore, the maximum (or the minimum) value of the function will be $f\left(\frac{-b}{2a}\right)$.

Calculation:

a.We can see that in $R(x)=9.5x-0.04x^2$, the coefficient of $x^2$ is $-0.04$. Therefore, we have $a=-0.04<0$.

Thus, the given function is having the maximum value at the vertex.

Thus, we get

$\frac{-b}{2a}=\frac{-(9.5)}{2(-0.04)}=118.75\approx 119$ And $f\left(\frac{-b}{2a}\right)=R(119)=9.5(119)-0.04{(119)}^2=564.06$

Thus, the vertex is at $(119,564.06)$.

Thus, the maximum revenue is $\$564.06$ and is obtained by selling 119 candy boxes.

b.$P(x)=R(x)-C(x)$

$=9.5x-0.04x^2-(1.25x+250)$

$=-0.04x^2+8.25x-250$

Thus, the profit function is $P(x)=-0.04x^2+8.25x-250$.

c.We can see that in $P(x)=-0.04x^2+8.25x-250$, the coefficient of $x^2$ is $-0.04$. Therefore, we have $a=-0.04<0$.

Thus, the given function is having the maximum value at the vertex.

Thus, we get

$\frac{-b}{2a}=\frac{-(8.25)}{2(-0.04)}=103.125\approx 103$ And $f\left(\frac{-b}{2a}\right)=P(103)=-0.04{(103)}^2+8.25(103)-250 = 175.39$

Thus, the vertex is at $(103,175.39)$.

Thus, the maximum profit is $\$1755.39$ and is obtained by selling 103 candy boxes.

d.The difference between parts (a) and (c) is that part (a) is the revenue function which is the total income that we get after selling the product whereas part (c) is of the profit function which is the gain we made by selling the product, i.e., profit is the gain that we get after subtracting the expenditure form the income.

Revenue is the total income that we get by selling a particular number of the product, that is $Revenue=no.products\times price/product$.

Let the number of products sold be $x$.

The price per product can be considered as a linear function as the price may increase or decrease with respect to time (or any other variable) and once it starts decreasing then it may not increase any more (in most circumstances).

Thus, let the price be $p(x)=a-bx$

Therefore, the revenue will be

$Revenue=no.products\times price/product$

$=x(a-bx)=ax-b{x}^2$

Thus, the revenue is a quadratic function.