Question

limited area with rectangles
A rectangular poster is to have an area of 200 square inches with a 1-inch margin on the sides and a 2-inch margin at the top and at the bottom. Find the dimensions of the poster with the largest printed area.

Answer 1

In this question, we need to maximize the printed area, let's call it \(A\).
We have the constraint, which is \(\large xy=200\), where \(x\) is the width of the paper, and \(y\) is the height of the paper;

Maximize: \(\large A = (x-2)(y-4)\)
Constraint: \(\large 200 = xy\)\[y=\frac{200}{x}\]Substitute this into the area function gives a function of x
\[\large A = (x-2)(\frac{200}{x}-4)\\A=(x-2)(\frac{200}{x}-\frac{4x}{x})=(x-2)(\frac{200-4x}{x})=(\frac{200-4x}{x})x-(\frac{200-4x}{x})2\\A=200-4x-(\frac{400-8x}{x})=200-4x-(\frac{400}{x}-\frac{8x}{x})=200-4x-\frac{400}{x}+8\\A=208-4x-\frac{400}{x}\]

Answer 2

Take the derivative of this and find its critical point

\[\large \frac{d}{dx}(A(x))=\frac{400}{x^2}-4\]

Answer 3

Set this equal to 0:\[0=\frac{400}{x^2}-4\\100=x^2\\x=\pm \sqrt{100}=\pm 10\]
A length could not be a negative number, so the width of this paper would be 10 inches, height of the paper 20 (200/10=20)

The printing area would be
Width: 10in - 2in = 8in
Height : 20in - 4in = 16in
8in*16in=128 inches squared

Answer 4

Thanks mang.

Answer 5

Good? :)

Answer 6

Excellent explanation. Thorough. I'm great :D

Answer 7

:)