Answer:
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.
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!!!