Chapter ISG, Problem 8.3.2P

a.

To determine

Explanation of Solution

Given:

Height of triangle= h

Breadth of rectangle= $5-2h$

Length of rectangle= $4h+2$

Formula used:

Area of triangle= $\frac{1}{2}\times base\times height$

Area of rectangle= $length\times breadth$

Calculation:

Area of figure= 2(area of triangle)+area of rectangle.

Rectangle,

Substituting values,

Area= $(5-2h)(4h+2)$

Solving,

Area= $5(4h+2)-2h(4h+2)$

Now,

Area= $20h+10-8h^2-4h$

Hence area of rectangle= $-8h^2+16h+10$

Triangle,

Here,

2( base ) = length of triangle

So,

  $base$ = $2h+1$

Substituting values in formula,

Area= $\frac{1}{2}\times (2h+1)\times h$

Solving,

Area= $\frac{(2h^2+h)}{2}$

Now, put values of area in above,

Area of figure= $2\times \frac{(2h^2+h)}{2}+\{-8h^2+16h+10\}$

Solving,

= $2h^2+h-8h^2+16h+10$

Adding terms,

= $-6h^2+17h+10$

Hence, this is the area of the figure.

b.

To determine

To find the maximum value of h

Answer to Problem 8.3.2P

  $h=\frac{10}{3}$

Explanation of Solution

Given:

Equation is $-6h^2+17h+10$

Calculation:

To find the maximum value of h,

Put $-6h^2+17h+10$ =0

Multiplying by -1,

  $6h^2-17h-10=0$

Grouping values,

  $6h^2-20h+3h-10=0$

Now taking common,

  $2h(3h-10)+1(3h-10)=0$

Taking common again,

  $(2h+1)(3h-10)=0$

For equation to be 0,

  $3h-10=0$ , $2h+1=0$

Taking

  $3h-10=0$ because $2h+1=0$ is negative and dimension cannot be in negative.

  $3h=10$

So,

  $h=\frac{10}{3}$

  $h=\frac{10}{3}$ is the maximum value of h for the maximum area.

Put value of h in equation.

Maximum area is 55.55 cubic unit