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!!!

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the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05

stethere all not eveanp-by-step explanation: