Question

56. A rectangular frame of uniform depth for a shadow box is to be made from a 36-in. piece of wood.

Answer 1

Answer and explanation:

Given : A rectangular frame of uniform depth for a shadow box is to be made from a 36-in. piece of wood.

To find :

a) Write a function to represent the display area in terms of x.

The perimeter of the box is 36 inches.

Let x and y be the length and width of the box.

So, $2x+2y=36$

$2y=36-2x$

$y=\frac{36-2x}{2}$

$y=18-x$ ....(1)

The area of the box is $A=lb$

$A=xy$

Substitute y from (1),

$A=x(18-x)$

$A=18x-x^2$

$A=-x^2+18x$

The function in terms of x, $A(x)=18x-x^2$

b. What dimensions should be used to maximize the display area?

As A is a quadratic function with a negative leading coefficient.

The vertex is the maximum point on the function.

So, the x-coordinate of the vertex is the value of x that will maximize the area.

$x=\frac{-b}{2a}$

Here, a=-1 and b=18

$x=\frac{-18}{2(-1)}$

$x=9$

Substitute in (1),

$y=18-9$

$y=9$

The dimension 9 in. by 9 in. should be used maximize the display area.

c) What is the maximum area?

Substitute the value of x in the area,

$A(x)=-x^2+18x$

$A(9)=-9^2+18(9)$

$A(9)=-81+162$

$A(9)=81$

The maximum area is 81 square inches.