A billboard designer has decided that a sign advertising the movie Fight Club II - Rage of the Mathematicians should have 1-ft. margins at the top and bottom, and 2-ft. margins on the left and right sides. Furthermore, the billboard should have a total area of 200 ft2 , including the margins. Find the dimensions that would maximize printed area.

Let 'X' be the total width of the billboard and 'Y' the total height of the billboard. The total area and printed area (excluding margins) are, respectively:

[tex]200 = x*y\\A_p = (x-4)*(y-2)[/tex]

Replacing the total area equation into the printed area equation, gives as an expression for the printed area as a function of 'X':

[tex]y=\frac{200}{x} \\A_p = (x-4)*(\frac{200}{x} -2)\\A_p=208 -2x -\frac{800}{x}[/tex]

Finding the point at which the derivate for this expression is zero gives us the value of 'x' that maximizes the printed area:

[tex]\frac{dA_p(x)}{dx} =\frac{d(208 -2x -\frac{800}{x})}{dx}=0\\0=-2 +\frac{800}{x^2} \\x=\sqrt{400}\\x=20\ ft[/tex]

If x = 20 ft, then y=200/20. Y= 10 ft.

The dimensions that maximize the printed area are:

Length = 20 ft

Height = 10 ft

MrRoyal MrRoyal · Answer 2 · 2021-10-29T00:00:53+02:00

The area of a shape is the amount of space it occupies

The dimension that would maximize the printed area is 10 by 20 ft.

Let the dimension of the billboard be represented with x and y.

So, we have:

[tex]\mathbf{Area=x \times y}[/tex]

The area is 200.

So, we have:

[tex]\mathbf{x \times y = 200}[/tex]

Make y the subject

[tex]\mathbf{y =\frac{200}{x}}[/tex]

When the margins are removed, the printed area is:

[tex]\mathbf{A = (x -2) \times (y - 4)}[/tex]

Substitute [tex]\mathbf{y =\frac{200}{x}}[/tex]

[tex]\mathbf{A = (x -2) \times (\frac{200}{x} - 4)}[/tex]

[tex]\mathbf{A = (x -2) \times (200x^{-1} - 4)}[/tex]

Expand

[tex]\mathbf{A = 200 - 4x - 400x^{-1} + 8}[/tex]

[tex]\mathbf{A = 208 - 4x - 400x^{-1} }[/tex]

Differentiate

[tex]\mathbf{A' = - 4 + 400x^{-2} }[/tex]

Set to 0

[tex]\mathbf{- 4 + 400x^{-2} = 0}[/tex]

Add 4 to both sides

[tex]\mathbf{ 400x^{-2} = 4}[/tex]

Rewrite as:

[tex]\mathbf{\frac{400}{x^2} =4}[/tex]

Inverse both sides

[tex]\mathbf{\frac{x^2}{400} =\frac 14}[/tex]

Multiply both sides by 400

[tex]\mathbf{x^2 =100}[/tex]

Take square roots

[tex]\mathbf{x =10}[/tex]

Recall that:[tex]\mathbf{y =\frac{200}{x}}[/tex]

So, we have:

[tex]\mathbf{y=\frac{200}{10}}[/tex]

[tex]\mathbf{y=20}[/tex]

The dimension that would maximize the printed area is 10 by 20 ft.

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