Chapter ISG, Problem 9.9.2.4P

a.

To determine

To write equation of area of gravel border.

Answer to Problem 9.9.2.4P

  $A(x)=56x+8x^2$

Explanation of Solution

Given information:

Length of rectangle is $12ft$

Width of rectangle is $8ft$

Let x be the width of the gravel border along the $12ft$ sides of the flower bed

Let 2x be the width of the gravel border along the $8ft$ sides of the flower bed

Formula used:

  $A=lb$

l is length of rectangle

b is width of rectangle

Calculation:

Length of rectangle $=12ft$

Width of rectangle $=8ft$

Area of flower bed

  $A=96f{t}^2$

Length of rectangle with gravel border $=(12+2x)$

Width of rectangle with gravel border $=(8+4x)$

Area of flower bed with gravel border

  $A=(12+2x)(8+4x)$

On solving,

  $A=(12)(8+4x)+(2x)(8+4x)$

On solving,

  $A=96+32x+24x+8x^2$

Function of area gravel border

  $A(x)=96+32x+24x+8x^2-96$

On solving,

  $A(x)=56x+8x^2$

b.

To determine

To the quadratic equation.

Answer to Problem 9.9.2.4P

Agreed

Explanation of Solution

Given information:

  $A(x)=56x+8x^2$

  $A(x)=56x+8x^2$

  $a>0$

Minimum value

We can identify the minimum or maximum value of a parabola by identifying the y -coordinate of the vertex.It can use a graph to identify the vertex or it can find the minimum or maximum value algebraically by using the formula $x=\frac{-b}{2a}$ . This formula will give the x -coordinate of the vertex.

  $x=\frac{-56}{2(8)}$

On solving,

  $x=\frac{-7}{2}$

Minimum point $x=\frac{-7}{2}$

  $A(\frac{-7}{2})=56(\frac{-7}{2})+8{(\frac{-7}{2})}^2$

On solving,

  $A(\frac{-7}{2})=-28\times 7+8\times \frac{49}{4}$

On solving,

  $A(\frac{-7}{2})=-98$

When x increases , the area of gravel border increases

So, agreed.