Chapter 2.7, Problem 36E

Solution

a.

To find: The area of page in terms of width $x$ .

The area of page in terms of width $x$ is $A=\left(2.5+\frac{40}{x}\right)\left(1.75+x\right)$ .

Given:

The print area is $40{\text{ in}}^{\text{2}}$ , left border is $0.75\text{ in}$ ,right border is $1\text{ in}$ ,top border is $1.5\text{ in}$ ,bottom border is $1\text{ in}$ .

Formula used:

The area of rectangle is $A=\text{Length}\times \text{Breadth}$

Calculation:

Let the width of print area be $x$ .

The area is length multiplied by breadth.

  $\begin{array}{c}\text{Length}\times \text{Width = 40}\\ \text{Length =}\frac{40}{x}\end{array}$

The total length of the page is top border plus bottom border plus the length of print area.

  $\begin{array}{c}L=1.5+1+\frac{40}{x}\\ =2.5+\frac{40}{x}\end{array}$

The total width of the page is left border plus right border plus the width of print area.

  $\begin{array}{c}W=0.75+1+x\\ =1.75+x\end{array}$

Conclusion:

The area of the page is $A=\left(2.5+\frac{40}{x}\right)\left(1.75+x\right)$

b.

To find: The width of the print area that has least area and also finds that least area.

The width of the print area is $31.62$ and the least area is $1268.98$ .

Given:

The print area is $40{\text{ in}}^2$ , left border is $0.75\text{ in}$ , right border is $1\text{ in}$ ,top border is $1.5\text{ in}$ ,bottom border is $1\text{ in}$ .

Formula used:

  $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=f\left(x\right)\frac{d}{dx}\left(g\left(x\right)\right)+g\left(x\right)\frac{d}{dx}\left(f\left(x\right)\right)$

  $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Calculation:

The area of page is $A=\left(2.5+\frac{40}{x}\right)\left(1.75+x\right)$ .

Differentiate the area $A=\left(2.5+\frac{40}{x}\right)\left(1.75+x\right)$ with respect to $x$ and equate to zero for the least area.

  $\begin{array}{c}\frac{dA}{dx}=\left(2.5+\frac{40}{x}\right)\frac{d}{dx}\left(1.75+x\right)+\left(1.75+x\right)\frac{d}{dx}\left(2.5+\frac{40}{x}\right)\\ \left(2.5+\frac{40}{x}\right)+\left(1.75+x\right)\left(-\frac{40}{x^2}\right)=0\\ -\frac{70}{x^2}-\frac{40}{x}=-2.5-\frac{40}{x}\\ -\frac{70}{x^2}=-2.5\end{array}$

Solve further for the value of $x$ .

  $\begin{array}{c}{x}^2=\frac{70}{2.5}\\ x=\sqrt{28}\\ x=5.29\end{array}$

Substitute $5.29$ for $x$ in the function $A=\left(2.5+\frac{40}{x}\right)\left(1.75+x\right)$ .

  $\begin{array}{c}A=\left(2.5+\frac{40}{5.29}\right)\left(1.75+5.29\right)\\ =\left(7.04\right)\left(10.06\right)\\ =70.83\end{array}$

Conclusion:

The width of the print area is $31.62$ and the least area is $1268.98$ .