Question

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

36 in.


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


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


c. What is the maximum area?

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=(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=(-b)/(2a)`

Here, a=-1 and b=18

`x=(-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.