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Q49.

Expert-verified

Algebra 2

Found in: Page 292

Book edition Middle English Edition

Author(s) Carter

Pages 804 pages

ISBN 9780079039903

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Short Answer

Steve has 120 feet of fence to make a rectangular kennel for his dogs. He will use his house as one side. What dimensions produce a kennel with the greatest area?

The dimensions that produce maximum area should be 30 ft and 60 ft.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information.

Steve has 120 feet of fence. He uses his house as one side.

Let

w = width of house

l = length of kennel

Step 2. Use the concept.

The y-intercept is the point where the graph intersects the y-axis.

Consider the function $f (x) = a x^{2} + b x + c, a \neq 0$ , the x-coordinate of vertex is $\frac{- b}{2 a}$ .

The graph of $f (x) = a x^{2} + b x + c, a \neq 0$

opens up and has a minimum value when $a > 0$ , and
opens down and has a maximum value when $a < 0$

Step 3. Solution.

The length of kennel is length + length + width = 120

Area of kennel is $length \cdot width$

So,

$\begin{array}{l} 2 l + w = 120 \\ \Rightarrow w = 120 - 2 l ............ (1) \end{array}$

Substitute the value of w in the equation of area:

$\begin{array}{l} A = l \cdot w \\ \Rightarrow A = l \cdot (120 - 2 l) ........ [w = 120 - 2 l] \\ \Rightarrow A = 120 l - 2 l^{2} \end{array}$

Writing this in the form of quadratic equation and optimizing:

$f (x) = - 2 x^{2} + 120 x$ , we have x $a = - 2, b = 120, c = 0$

Here, $a = - 2 < 0$

So, the graph opens down and has a maximum value.

The maximum value of the function is the y-coordinate of the vertex and is the maximum height reached by the object.

The x-coordinate of the vertex is

. $\begin{array}{l} \frac{- b}{2 a} \\ = \frac{- (120)}{2 (- 2)} .......... [a = - 2, b = 120] \\ = 30 \end{array}$

So, length of kennel is 30.

Substitute $l = 30$ in equation (1):

$\begin{array}{l} w = 120 - 2 l \\ \Rightarrow w = 120 - 2 (30) \\ \Rightarrow w = 120 - 60 \\ \Rightarrow w = 60 \end{array}$

So the width of kennel should be 60 ft and two sides must be 30 ft, and then his house would be 60 ft to accommodate the 4th side of the kennel.

The dimensions that produce maximum area should be 30 ft and 60 ft.

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