Answer:
Length = 20 ft
Height = 10 ft
Step-by-step explanation:
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
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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