A rectangular mural will have a total area of 24 ft, which includes a boarder of 1 ft on the left right and bottom and a border of 2 on the top. what dimensions maximize the total paintable area inside the boarder

The dimensions of the mural that maximize the total paintable area inside the border = 4 ft × 6 ft

Making the dimensions of the maximum paintable area to be 2 ft × 3 ft.

Step-by-step explanation:

Let the dimensions of the rectangular mural be x and y respectively.

The total area of the mural = 24 ft²

xy = 24

Then there's a 1 ft border on the left, right and bottom.

A 2 ft border at the top.

The dimensions of the printable area will be (x-2) ft and (y-3) ft

The area of the printable area will then be

(x-2)(y-3) = A(x,y) = (xy - 3x - 2y + 6)

This is the area to be maximized

Recall xy = 24; y = (24/x)

We can substitute for y in the area to be maximized

A(x,y) = (x-2)(y-3) = (xy - 3x - 2y + 6)

A(x) = x(24/x) - 3x - 2(24/x) + 6

A(x) = 24 - 3x - (48/x) + 6 = 30 - 3x - (48/x)

We then have to maximize

A(x) = 30 - 3x - (48/x)

First of, we obtain (dA/dx)

(dA/dx) = -3 + (48/x²)

At maximum point, (dA/dx) = 0

-3 + (48/x²) = 0

3x² = 48

x² = 16

x = 4 or -4

For the maximum of the function of the area, x = 4, y = (24/4) = 6

The area of the paintable part of the mural

A(x,y) = (x-2)(y-3)

At maximum area, x = 4 ft, y = 6 ft

Maximum paintable area = (4-2)(6-3) = (2)(3) = 6 ft²

The dimensions maximize the total paintable area inside the border = 4 ft × 6 ft

Hope this Helps!!!