Chapter ISG, Problem 8.2.3P

a.

To determine

To find: The expression of the area of trapezium.

  

Answer to Problem 8.2.3P

Expression of Area is $8h-h^2$ .

Explanation of Solution

Given information:

The given trapezium is,

  

Calculation:

General formula for trapezium is given by,

  $A=\frac{1}{2}\times \left(\text{Sum of lenght of parallel line}\right)\times \left(\text{perpendicular height between it}\right)$

Sum of parallel lines,

  $\begin{array}{c}a+b=14-3h+h+2\\ =16-2h\end{array}$

Perpendicular height between it is, $h$ .

  $\begin{array}{c}A=\frac{1}{2}\left(16-2h\right)\times h\\ =8h-h^2\end{array}$

Expression of Area is $8h-h^2$ .

b.

To determine

To find: The maximum area of trapezoid and the height that gives this area.

Answer to Problem 8.2.3P

The maximum area is $12\text{sq unit}$ and height is $2\text{units}$ .

Explanation of Solution

Given information:

Expression of Area is $8h-h^2$

Calculation:

The given area is,

  $A=8h-h^2$ …..(1)

Differentiate equation 1 with respect to h .

  $\begin{array}{c}\frac{dA}{dh}=\frac{d}{dh}\left(8h-h^2\right)\\ =8-2h\end{array}$ …..(2)

Equate equation 2 with respect to h .

  $\begin{array}{c}\frac{dA}{dh}=0\\ 8-2h=0\\ h=\frac{8}{2}\\ h=4\end{array}$

Again differentiate equation 2 with respect to h .

  $\begin{array}{c}\frac{d^{2}A}{d{h}^2}=\frac{d}{dh}\left(8-2h\right)\\ =-2<0\end{array}$

Since second derivative is less than 0, the maximum area will be at $h=2$ .

Maximum Area is,

  $\begin{array}{c}A=8\left(2\right)-{\left(2\right)}^2\\ =16-4\\ =12\end{array}$

Hence, the maximum area is $12\text{sq unit}$ and height is $2\text{units}$ .