Chapter 3.5, Problem 11CYU

a.

To determine

To find: Function $n\left(p\right)$ that represents the costs of paints with the $15\%$ discount.

Answer to Problem 11CYU

  $n\left(p\right)=5.9415p$

Explanation of Solution

Given information:

Cost of paints = $f\left(p\right)=6.99p$

There is a coupon of discount of $15\%$ on purchase of paint.

Cost of paints = $f\left(p\right)=6.99p$

Discount = $15\%$ of total purchase

  $\Rightarrow Discount=\left(\frac{15}{100}\right)\times 6.99p$

  $\begin{array}{l}\Rightarrow Discount=\left(0.15\right)\times 6.99p\\ \Rightarrow Discount=1.0485p\end{array}$

Therefore

Final Cost of paints = Cost of paints − Discount

  $\begin{array}{l}\Rightarrow n\left(p\right)=f\left(p\right)-1.0485p\\ \Rightarrow n\left(p\right)=6.99p-1.0485p\\ \Rightarrow n\left(p\right)=5.9415p\end{array}$

Therefore,

Cost of paints with the $15\%$ discount = $n\left(p\right)=5.9415p$

b.

To determine

To find:Cost of purchasing 5 paints before and after the discount.

Answer to Problem 11CYU

Cost of paints before discount = 34.95

Cost of paints after discount = 29.7075

Explanation of Solution

Given information:

Cost of paints = $f\left(p\right)=6.99p$

There is a coupon of discount of $15\%$ on purchase of paint.

Cost of paints before discount = $f\left(p\right)=6.99p$

Substitute $p=5$

We get

  $\begin{array}{l}f\left(p\right)=6.99p\\ \Rightarrow f\left(5\right)=6.99\times 5\\ \Rightarrow f\left(5\right)=34.95\end{array}$

Therefore,

Cost of paints before discount = 34.95

We got

Cost of paints after discount = $n\left(p\right)=5.9415p$

Substitute $p=5$

We get

  $\begin{array}{l}n\left(p\right)=5.9415p\\ \Rightarrow n\left(5\right)=5.9415\times 5\\ \Rightarrow n\left(p\right)=29.7075\end{array}$

Therefore,

Cost of paints after discount = 29.7075