Pre-Calc problem oo help
What is the smallest perimeter possible for a rectangle whose area is 16 in^2 and what are its demensions?
Here is what I did:
P=2x+2y
A=xy
(16/x)=y
P=2x+2(16/x)
P=2x+(32/x)
P'=2-(32/x^2)
2-(32/x^2)=0
x=4
I used the sign analysis and I found that x=4 is a local minimum.
Did I do everything right so far? o_o
May 3, 2013
6 Comments • Newest first
You know the original dimensions are 8 x 15 inches. You also know that the volume you would get from the fold is is going to be ( 8 in - 2*height)*(15 in - 2*height)*height since you are cutting squares from the corners and folding up. Just more optimization like last time from here, which you seem to grasp.
Edit: Changed it to 2*height like it should be, I wasn't paying enough attention sorry.
[quote=LowWillpower]Yes. It's also possible to see how your answer should just always be square root of the area for both sides when minimizing perimeter by noticing P'/2 = 1 - A/x^2 and since P'/2 is directly proportional to P', P'/2 = 0 means P' = 0 so 1 = A/x^2 and x = sqrt(A)[/quote]
I have another problem if you can help T.T
You are planning go make an open rectangular box from an 8in by 15in piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way and what is the volume?
[quote=SriLankanKen]So I did it right? o_o[/quote]
Yes. It's also possible to see how your answer should just always be square root of the area for both sides when minimizing perimeter by noticing P'/2 = 1 - A/x^2 and since P'/2 is directly proportional to P', P'/2 = 0 means P' = 0 so 1 = A/x^2 and x = sqrt(A)
[quote=chaos0]Just solve for a variable of the area equation, sub that variable into a variable for the perimeter equation. Then find the derivative of the equation and plug the given small x value into the derivative equation[/quote]
So I did it right? o_o
Yes, seems like you have. I know this because if there's one thing I remember from these questions, the answer is a square (and a side of length 4 is obviously the square root of 16).
Just solve for a variable of the area equation, sub that variable into a variable for the perimeter equation. Then find the derivative of the equation and plug the given small x value into the derivative equation