Chapter 1.5, Problem 5MRPS

Solution

a.

To write: A function that gives the pizza shop's revenue $R$ if there are $x$ price increases

A function that gives the pizza shop's revenue $R$ if there are $x$ price increases is $R\left(x\right)=\left(2+0.25x\right)\left(80-5x\right)$

Given:

A pizza shop sells about $80$ slices of pizza each day during lunch when it charges $\$2$ per slice. For each $\$.25$ increase in price, about $5$ fewer slices are sold each day during lunch.

Calculation:

After $x$ price increases, the total number of slices sold is $80-5x$ , and the price of each slice is $2+0.25x$

Shop's revenue is the product of the total sales and the price of each slice.

The function that gives the pizza shop's revenue is $R\left(x\right)=\left(2+0.25x\right)\left(80-5x\right)$ .

b.

To find: A value of $x$ maximizes $R$

A value of $x$ maximizes $R$ is $4$

Given:

A pizza shop sells about $80$ slices of pizza each day during lunch when it charges $\$2$ per slice. For each $\$.25$ increase in price, about $5$ fewer slices are sold each day during lunch.

Calculation:

In order to find the value of $x$ that maximizes $R$ , write the equation in intercept form. For this, first apply the FOIL method and simplify.

  $\begin{array}{c}R\left(x\right)=160-10x+20x-1.25x^2\\ =-1.25x^2+10x+160\end{array}$

Take out the common factor $-1.25$ .

  $R\left(x\right)=-1.25\left(x^2-8x-128\right)$

Factor the expression within the parentheses. Now, find two factors $m$ and $n$ such that their product gives $-128$ and sum gives $-8$ . Two such factors are $-16$ and $8$ .

Thus, $R\left(x\right)=-1.25\left(x-16\right)\left(x+8\right)$ .

Now, we have to find the $x$ -coordinate of the vertex. For a quadratic function of the form $y=a\left(x-p\right)\left(x-q\right)$ , the $x$ -coordinate of the vertex of its graph is $\frac{p+q}{2}$ .

In this case, $p$ is $16$ and $q$ is $-8$ . The $x$ -coordinate of the vertex is $\frac{-8+16}{2}$ or 4. Therefore, the value of $x$ that maximizes $R$ is $4$ . This means that the revenue is maximum when the price per slice is $\$\left[2+0.25\left(4\right)\right]$ or $\$3$ .

c.

To find: Reason behind negative $x-$ value in part (b).

Value of $x$ is negative because the major portion of the parabola lies on the negative $x$ -axis.

Given:

A pizza shop sells about $80$ slices of pizza each day during lunch when it charges $\$2$ per slice. For each $\$.25$ increase in price, about $5$ fewer slices are sold each day during lunch.

Explanation:

After $x$ price decreases, the total number of slices sold is $80+5x$ , and the price of each slice $2-0.25x$ .

The function that gives the pizza shop's revenue is $R\left(x\right)=\left(2-0.25x\right)\left(80+5x\right)$ .

Apply the FOIL method and simplify.

  $\begin{array}{c}R\left(x\right)=160+10x-20x-1.25x^2\\ =-1.25x^2-10x+160\end{array}$

Take out the common factor $-1.25$ .

  $R\left(x\right)=-1.25\left(x^2+8x-128\right)$

Factor the expression within the parentheses. Find two factors $m$ and $n$ such that their product gives $-128$ and sum gives $8$ . Two such factors are $16$ and $-8$ .

Thus, $R\left(x\right)=-1.25\left(x+16\right)\left(x-8\right)$ .

Now, find the $x$ -coordinate of the vertex. $\frac{p+q}{2}$ .

In this case, $p$ is $-16$ and $q$ is $8$ . The $x$ -coordinate of the vertex is $\frac{-16+8}{2}$ or $-4$ . Therefore, the value of $x$ that maximizes $R$ is $-4$ . Value of $x$ is negative because the major portion of the parabola lies on the negative $x$ -axis.