Optimization of Area of cardboard poster- need more help

[sofiaahmed]

q: A cardboard poster containing 512 inches squared of printed region is to have a margin of 2 inches at the top, 2 inches at the bottom, and 1 inch at each side. Determine the dimensions of a rectangular piece of cardboard with the smallest are that can be used to make such a poster.

I have set 512 as the constraint. The length of one side is (512/x) + 4 and the other one is x + 2. I multiplied the two lengths and took the derivative. The answer I got for x, approximately 15, doesn't make sense though when I plug it back in.

 

[tkhunny]

16 is approximately 15? You had better try the algebra again. Something about your solution went haywire. If you still have an 'x'-term, you missed something.

 

[sofiaahmed]

I don't know what you mean by 16, and I also did not have an x term left, I solved for x.

 

[tkhunny]

\(\displaystyle f(x)\;=\;(\frac{512}{x}\;+\;4)*(x\;+\;2)\)

\(\displaystyle f(x)\;=\;512\;+\;\frac{1024}{x}\;+\;4x\;+\;8\)

\(\displaystyle f'(x)\;=\;4*(\frac{x^{2}-256}{x^{2}})\) <== See, no 'x' term

Now what do you get for the value of x?

 

[soroban]

Re: Optimization of Area

Hello, Sofia!

A cardboard poster containing 512 in² of printed region
is to have a margin of 2 inches at the top, 2 inches at the bottom, and 1 inch at each side.
Determine the dimensions of a rectangular piece of cardboard
with the smallest area that can be used to make such a poster.

I have set 512 as the constraint.
The length of one side is \(\displaystyle \frac{512}{x}\,+\,4\) and the other one is \(\displaystyle x\,+\,2\)
I multiplied the two lengths and took the derivative. Correct!
The answer I got for x, approximately 15 . . . I got 16
doesn't make sense though when I plug it back in.

The area is: \(\displaystyle \:A\;=\;\left(\frac{512}{x}\,+\,4\right)(x\,+\,2) \;=\;512\,+\,\frac{1024}{x}\,+\,4x\,+\,8 \;=\;520\,+\,1024x^{-1}\,+\,4x\)

Differentiate and equate to zero: \(\displaystyle \:A'\;=\;-1024x^{-2}\,+\,4\;=\;0\)

Multiply by \(\displaystyle x^2:\;-1024 \,+\,4x^2\:=\:0\;\;\Rightarrow\;\;x^2\:=\:256\;\;\Rightarrow\;\;x\:=\:16\)


The width of the poster is: \(\displaystyle \,x\,+\,2\:=\:18\text{ inches.}\)

The height is: \(\displaystyle \,\frac{512}{x}\,+\,4\:=\:36\text{ inches.}\)

 

[sofiaahmed]

Thank you both for your replies. My only question is height times width (18x36) does not equal 512, am I reading the problem wrong? I think there is something I am not understanding. Thanks again.

 

[tkhunny]

You are not distinguishing between the entire poster (including borders) and the smaller printable area (512).