Calculus optimization question?

Question

A poster is to have an area of 180 square inches with 1 inch margins at the bottom and sides and a 2 inch margin at the top. What dimensions will give the largest printed area?

Answer 1

Let x and y be sides of entire poster.
Then xy = 180 (sq. inches)

Area = (x-2)(y-3) = xy -3x-2y+6
=xy-3x-2y+6
=180-3x-2(180/x)+6
=186-3x-360/x
d(Area)dx=-3+360/(x^2)
To maximize/minimize:
-3+360/(x^2) = 0
-3(x^2)+360=0
3(x^2)=360
x^2=120
so x = sqrt(120) or x = -sqrt(120)
x = 10.95 inches
So y = 16.43 inches
Dimensions are 10.95 by 16.43 inches.

Answer 2

The idea is to have a function you can optimize.
Here you would like to model the area of print as a function of the dimensions.

A = (printed length)*(printed width)

You know that the poster has an area of 180. Let x = poster width and y = poster length. So, x*y = 180. That's all well and good but we need to relate that to the printed length and width using the margins. The top and bottom together make 3 inches. So the printed length is y - 3. The sides have 1 inch each or 2 inches total. So the printed width is x - 2. Now we need to put y in terms of x. Since, x*y = 180, y = 180/x.
So, the printed length is (180/x) - 3.

So the printed area function is now:
A = ( (180/x) - 3)*(x - 2) = 180 - (360/x) - 3x + 6.

Differentiate: A' = (360/x^2) - 3

Set it equal to zero: 0 = (360/x^2) - 3.

Solve for x: 360/x^2 = 3, 360 = 3x^2, 120 = x^2, x = sqrt(120)
So, x is an irrational number - pretty darn close to 11.
If you want it as a simplified root its 2sqrt(30).
Then, y = 180/(2sqrt(30)) - 3 = 3sqrt(30) - 3.

So the poster is 3sqrt(30) - 3 by 2sqrt(30).

Answer 3

The printed area is the portion of the 180 inches^2 without the borders.

Area of the entire poster is A = xy, with x as the width and y as the length.

To find the area of the printed area, subtract the borders.

The sides have a 1 inch border, so, the inner width is x - 2
The top has a 2 inch border and the bottom a 1 inch border, so the inner length is y - 3
Area of printed portion = P = (x - 2)(y - 3)

But with 180 = xy, y = 180/x

So, P = (x - 2)( (180/x) - 3) = 180 - (360/x) - 3x + 6
Maximizing printed area, take the derivative of P
and set equal to zero

P' = -3 + (360/x^2) = 0
(360/x^2) = 3
3x^2 = 360
x^2 = 120

x = 11 inches
y = 180/11 = 16.4inches

Answer 4

Length of posture = x
Breadth of posture = y
Area xy=180, and therefore y=180/x ------(1)

Print length = x-2
Print breadth = y-(2+1)
= y-3
Print area = (x-2)(y-3) = A
i.e., xy-3x-2y+6=A
Therefore A=186-3x-360/x [Substituting (1)
(dA/dx) = -3 + 360/ squared x
and second order diffn is -360 / cubic x which is negative.
Therefore A is maximum when (dA/dx) = -3 + 360/ squared x =0
Solving this we get x = Square root of 120 and therefore y=180/Square root of 120 are the required dimensions of the poster to give the largest printed area.