Here is a problem that I think I know how to solve, but can't get the right answer.  Am I on the right track?

sum of the perimeters of a square and an equaliteral triangle is 10. find the dimensions that will maximize the area.

Let y= lenght of the side of the triangle, x= length of the side of the square 

3y+4x=10; from here y=(10-4x)/3

A(x)=x^2+ (3) y^2/4; replace y with (10-4x)/3

Find A'(x) and see when it's equal to zero. One of the values of x will give me a dimension that maximizes the area; from there, I can get the other dimension.

Does it look OK or am I missing something here?  Anna.

You have it exactly right.  Nice work.

                      --CFW

I computed the derivative of

x ²  + 3((10-4x)/3) ² /4 to be after simplifying

2x + 23((4 - 10x))/9 which when set equal to 0 yields x = 103/(9 + 43).

Of course, as is the case with all of my calculations, this should be checked.

At any rate, the set up that you described is correct.

              --CFW