Chapter 1, Problem 3TP

Solution

To find: The largest possible frame width.

The largest possible frame width $x=3.6$ inches.

Given information

Given figure is:

  

Calculation:

The width of the frame= $x$

The side of photo =10

The length and breadth of outer frame is= $10+2x$

The area of photo= $side\times side$

The area of photo= $10\times 10=100$

Area of the frame= $\left(10+2x\right)\left(10+2x\right)$

  $\begin{array}{l}A=\left(10+2x\right)\left(10+2x\right)\\ A=100+4x^2+40x\end{array}$

  $\text{Area of the frame =Total area covered by the tiles}$

  $100+4x^2+40x=300$

  $\begin{array}{l}A=4x^2+40x-200=0\\ A=x^2+10x-50=0\end{array}$

Solve the quadratic equation

  $\begin{array}{l}{x}^2+10x-50=0\\ x=\frac{-10\pm \sqrt{{\left(-10\right)}^2-4\left(-50\right)}}{2}\end{array}$

  $\begin{array}{l}x=\frac{-10\pm \sqrt{100+200}}{2}\\ x=\frac{-10\pm 10\sqrt{3}}{2}\\ x=-5+5\sqrt{3},-5-5\sqrt{3}\end{array}$

  $\begin{array}{l}x=-5+5\cdot 1.73\\ x=-5+8.65\\ x=3.65\end{array}$

Find $\frac{d^{2}A}{d{x}^2}$

  $\frac{d^{2}A}{d{x}^2}=2>0$

Hence the area of the frame will be largest when width $x=3.6$ inches.