Chapter 3.3, Problem 65E

(a)

To determine

To find: The maximum weekly revenue.

Answer to Problem 65E

  $\$3600$

Explanation of Solution

Given: The shop’s revenue can be modeled by $y=\left(70-x\right)\left(50+x\right)$

The model of weekly revenue of shop, $y=\left(70-x\right)\left(50+x\right)$

Using intercept form to find maximum revenue.

For x -intercept, set each factor to 0 and solve for x.

  $\therefore x=-50,70$

The mid value of x -intercept attain maximum revenue.

  $\therefore \text{Revenue maximum at }x=\frac{70-50}{2}=10$

  $y_{\max }=\left(70-10\right)\left(50+10\right)=3600$

(b)

To determine

To write: The function in vertex form and find maximum weekly revenue.

Answer to Problem 65E

  $\begin{array}{l}y=-{\left(x-10\right)}^2+3600\\ y_{\max }=3600\end{array}$

Explanation of Solution

Given: The shop’s revenue can be modeled by $y=\left(70-x\right)\left(50+x\right)$

The given modeled of revenue, $y=\left(70-x\right)\left(50+x\right)$

Now change into vertex form like, $y=a{\left(x-h\right)}^2+k$

  $\begin{array}{l}y=\left(70-x\right)\left(50+x\right)\\ y=-x^2+20x+3500\\ y=-\left(x^2-20x\right)+3500\left[\text{Completing square}\right]\\ y=-{\left(x-10\right)}^2+3600\left[\text{Vertex form}\right]\end{array}$

Maximum weekly revenue at vertex $x=10$

  $y_{\max }=-{\left(10-10\right)}^2+3600=3600$

(b)

To determine

To choose: The best method to find the maximum revenue.

Answer to Problem 65E

Vertex form

Explanation of Solution

Given: The shop’s revenue can be modeled by $y=\left(70-x\right)\left(50+x\right)$

Using vertex form very easy to find the maximum. Because it given vertex directly and maximum of parabolic curve at vertex.