Chapter 11.2, Problem 41PPS

(a)

To determine

To Find: The domain and range of the function.

Answer to Problem 41PPS

Domain: All real number greater than $0$

Range: All real numbers between $0$ to $16$

Explanation of Solution

Given:

The equation $h=\frac{2(64)}{b_1+8}$ represents the height $h$ of a quadrilateral with an area of $64$ square units. The quadrilateral has two opposite sides that area parallel and $h$ units apart, one is $b_1$ units long and other is $8$ units long.

Given: $h=\frac{2(64)}{b_1+8}$

Since the side of the quadrilateral cannot be negative

Thus

Domain: All real number greater than $0$

Range: All real numbers between $0$ to $16$

(b)

To determine

To Graph: The function in first quadrant.

Answer to Problem 41PPS

  

Explanation of Solution

Given:

The equation $h=\frac{2(64)}{b_1+8}$ represents the height $h$ of a quadrilateral with an area of $64$ square units. The quadrilateral has two opposite sides that area parallel and $h$ units apart, one is

  $b_1$ units long and other is $8$ units long.

Given: $h=\frac{2(64)}{b_1+8}$

The graph of the function is

  

(c)

To determine

To Estimate: The value of $h$ when $b_1=10$

Answer to Problem 41PPS

  $h=7.11$

Explanation of Solution

Given:

The equation $h=\frac{2(64)}{b_1+8}$ represents the height $h$ of a quadrilateral with an area of $64$ square units. The quadrilateral has two opposite sides that area parallel and $h$ units apart, one is $b_1$ units long and other is $8$ units long.

Given: $h=\frac{2(64)}{b_1+8}$

  

Using graph it is seen that

When $b_1=10$ , $h=7.11$