Mimimum&Maximum problem. HelpASAP(:

[NikkiGurl.]

A wire 17 inches long is cut into 2 pieces. One of these sections is bent to form a square. The other section is bent to form a rectangle in which is twice as long as its width. How should the original wire be cut so that the sum of the area of the square and the rectangle is a minimum?

Well, I cannot figure this out for the life of me. And it's quite frustrating. The 2 pieces are not equal, so please don't assume they are.
I have started with 17=a+b since I'm finding a sum.
I know that the equation for the perimeter of a square is 4s=P so ive used 4a in place of a.
Also the equation for the perimeter of a rectangle is 2L+2W=P , so ive used 2(b+2b) in place of b.
So I have (4a)+(2(b+2b))=17 ?
So I isolate a variable and then substitute from here?
Or am I totallyy off ?
Please helpp!

 

[Loren]

A wire 17 inches long is cut into 2 pieces. One of these sections is bent to form a square. The other section is bent to form a rectangle in which is twice as long as its width. How should the original wire be cut so that the sum of the area of the square and the rectangle is a minimum?

Well, I cannot figure this out for the life of me. And it's quite frustrating. The 2 pieces are not equal, so please don't assume they are.
I have no idea how to even start this problem..All I know is I'm supposed to get an equation some how to put into the calculator to figure out the minimum point.. But how do I do that??
Please Help ASAP!

Start by naming things. You know that if you cut a 17 inch wire into two pieces and one of them is named x, then the other one is named 17-x. With those two pieces, write the equations for the areas of each of the figures. If the piece that is x inches long is formed into a square, then the area of the square is (1/4*x)[sup:2ny0d54z]2[/sup:2ny0d54z]. You have some information about the sides of the rectangle so do a similar thing with that. Then, you find the derivative of the sum of the areas, etc. It might be a little less cumbersome if you use 17-x piece for the square and the x piece for the rectangle.

 

[NikkiGurl.]

Okay wait,
why did you put(1/4*x)^2
Where did you get the 1/4 from?
I completly dont understand what your saying..
Sorry.
Can you please explain?

 

[BigGlenntheHeavy]

\(\displaystyle We \ have \ a \ square \ with \ s \ sides \ and \ a \ rectangle \ with \ two \ sides \ of \ x \ (width) \ and \ two \ sides\)\(\displaystyle of \ 2x \ (length).\)

\(\displaystyle Hence, \ 4s \ + \ 6x \ = \ 17, \ (perimeter \ of \ both \ polygons).\)

\(\displaystyle A \ = \ s^{2} \ + \ 2x^{2}, \ (area \ of \ both \ polygons).\)

\(\displaystyle Now, \ 4s \ = \ 17-6x, \ s \ = \ \frac{17-6x}{4}.\)

\(\displaystyle Hence, \ total \ area \ of \ the \ two \ polygons \ = \ A \ = \ \bigg(\frac{17-6x}{4}\bigg)^{2}+2x^{2}.\)

\(\displaystyle Now, \ find \ \frac{dA}{dx}, set \ to \ zero \ and \ solve \ for \ x. \ Then \ solve \ for \ s.\)

\(\displaystyle Can \ you \ take \ it \ from \ here?\)

 

[NikkiGurl.]

\(\displaystyle We \ have \ a \ square \ with \ s \ sides \ and \ a \ rectangle \ with \ two \ sides \ of \ x \ (width) \ and \ two \ sides\)\(\displaystyle of \ 2x \ (length).\)

\(\displaystyle Hence, \ 4s \ + \ 6x \ = \ 17, \ (perimeter \ of \ both \ polygons).\)

\(\displaystyle A \ = \ s^{2} \ + \ 2x^{2}, \ (area \ of \ both \ polygons).\)

\(\displaystyle Now, \ 4s \ = \ 17-6x, \ s \ = \ \frac{17-6x}{4}.\)

\(\displaystyle Hence, \ total \ area \ of \ the \ two \ polygons \ = \ A \ = \ \bigg(\frac{17-6x}{4}\bigg)^{2}+2x^{2}.\)

\(\displaystyle Now, \ find \ \frac{dA}{dx}, set \ to \ zero \ and \ solve \ for \ x. \ Then \ solve \ for \ s.\)

\(\displaystyle Can \ you \ take \ it \ from \ here?\)

OHMYGOSH!
I completly understand it! Thank you so much!