Respuesta :
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, [tex]2x+2y=36[/tex]
[tex]2y=36-2x[/tex]
[tex]y=\frac{36-2x}{2}[/tex]
[tex]y=18-x[/tex] ....(1)
The area of the box is [tex]A=lb[/tex]
[tex]A=xy[/tex]
Substitute y from (1),
[tex]A=x(18-x)[/tex]
[tex]A=18x-x^2[/tex]
[tex]A=-x^2+18x[/tex]
The function in terms of x, [tex]A(x)=18x-x^2[/tex]
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.
[tex]x=\frac{-b}{2a}[/tex]
Here, a=-1 and b=18
[tex]x=\frac{-18}{2(-1)}[/tex]
[tex]x=9[/tex]
Substitute in (1),
[tex]y=18-9[/tex]
[tex]y=9[/tex]
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,
[tex]A(x)=-x^2+18x[/tex]
[tex]A(9)=-9^2+18(9)[/tex]
[tex]A(9)=-81+162[/tex]
[tex]A(9)=81[/tex]
The maximum area is 81 square inches.
