Chapter 13.6, Problem 20PSD

To determine

To find: The maximum value of the m.

Answer to Problem 20PSD

The maximum value of $m=\frac{-8}{21}$ .

Explanation of Solution

Given:

The quadrilateral region that is bounded by the graph of the equation $y=mx+3$ , here $x=2,5$ and $y=1$ the area is 2.

Calculation:

Consider the required diagram is shown in Figure 1

  

Figure 1

Then, the area of the trap is,

  $\begin{array}{l}A=\frac{h}{2}\left(b_1+b_2\right)\\ 2=\frac{3}{2}\left(2m+2+5\text{m}+2\right)\\ m=\frac{-8}{21}\end{array}$