I believe you'll find that this extreme point is a minimum not a maximum. Thus your maximum lies at a boundary of your constraint equation.
You should try plotting the Area and y as a function of x to get a better idea of what is going on.
The sum of the perimeter of a equilateral triangle and square is 10. Find the dimensions of the triangle of and square that produce a minimum total area.
My Work:
Perimeter of square is 4x.
Perimeter of Equilateral is 3y.
Area of square is x^2.
Area of Equilateral triangle is [y^2*sqrt(3)]/4.
Total area is x^2 + [y^2*sqrt(3)/4].
Total perimeter is 3y + 3x, where 3y + 4x = 10.
I then solved 10 = 3y + 4x for y and got y = (10-4x)/3.
Afterward, I plugged y into the total perimeter.
Let T = total perimeter
T = x^2 + (1/4)[(10-4x)/3]^2*sqrt(3).
I then found T prime.
T' = 2x + (2x-5)/3*sqrt(3)
Let T' = 0 to find x.
I found out that x = 5/[6*sqrt(3) + 2]
I stopped here because it does not make sense to me.
Can someone get me started in the right direction?
I believe you'll find that this extreme point is a minimum not a maximum. Thus your maximum lies at a boundary of your constraint equation.
You should try plotting the Area and y as a function of x to get a better idea of what is going on.
So far so good.
This is actually a formula for area, not perimeter. But that's actually what you want, so as long as we remember that T=area we can continue:
This is not right. I get: T' = 2x + (4/9) sqrt(3)(2x-5)
If you continue from here you'll find a value for x where T' is zero, But then you must check to see whether this gives a min or a max. Since T is a quadratic in x, you can see that the second derivative of T will be positive throughtout the domain, so the point we just calculated is a min, not a max. Hence as romsek noted - the max value for T will occur at the boundary of the domain, where either x=0 or y= 0.