Chapter 4.4, Problem 5E

a.

To determine

To find: To find the y- co ordinate point.

Answer to Problem 5E

The y-co ordinate of point P is $-x+1$ .

Explanation of Solution

Given:

The function $x+y=20$ .

Calculation:

Given

  

Given the hypotenuse of an isosceles triangle is $2$ unit longer.

Let the numbers are $x$ and $y$

  $0<x<20$ .

  $\begin{array}{l}x+y=20\\ y=20-x\end{array}$

The equation of straight-line AB is

  $y=-x+1$

The y- coordinates of point p is $-x+1$ .

b.

To determine

To find: To find the area of the rectangle.

Answer to Problem 5E

The Area of the rectangle is $-2x^2+2x$ .

Explanation of Solution

Given:

The sides are $-x+1$ and $x$

Calculation:

Area of the rectangle

  $\begin{array}{l}A(x)=2x(-x+1)\\ A(x)=-2x^2+2x\end{array}$

c.

To determine

To find: To find the maximum area and dimensions.

Answer to Problem 5E

Maximum area is $\frac{1}{2}$ and dimensions are $\frac{1}{2},1$ .

Explanation of Solution

Given:

The function $A=-2x^2+2x$

Calculation:

  $\frac{dA}{dx}=-4x+2$

For any function $f(x)$ the condition of minimum and maximum is obtained by finding $f\text{'}(x)=0$ critical points and end points.

Critical points occur at,

  $\begin{array}{l}A\text{'}=0\\ -4x+2=0\\ x=\frac{2}{4}\\ x=\frac{1}{2}\end{array}$

Solving it graphically, we get it has critical point occurs at $x=\frac{1}{2}$ .

  $\frac{dA}{dx}>0$ for $0<x<\frac{1}{2}$ and $\frac{dA}{dx}<0$ for $\frac{1}{2}<x<1$ this corresponds to maximum area

For,

  $\begin{array}{l}x=\frac{1}{2}\\ y=2x=1\\ A=-2x^2+2x\\ A=\frac{1}{2}\end{array}$

Thus, for maximum area the rectangle is $\frac{1}{2}{\text{ m}}^2$ and its dimensions are $\frac{1}{2},1$ .