The sum of the perimeters of an equilateral triangle and a square is 10. Find the dimensions of the triangle and square that produce a minimum area.

I would like to solve it by incorporating A=(sqrt(3)x^2)/4 (area of an equilateral triangle)

Question

The sum of the perimeters of an equilateral triangle and a square is 10. Find the dimensions of the triangle and square that produce a minimum area.

I would like to solve it by incorporating A=(sqrt(3)x^2)/4  (area of an equilateral triangle)

anonymous · Answer

put the side of the triangle as $x$ and the side of the square is $y$ then 
$$3x+4y=10$$ solve for $y$ get $$y=\frac{10-3x}{4}$$

anonymous · Answer

total area is therefore  what you said 
triangle area is $A(x)=\frac{\sqrt{3}x^2}{4}+(\frac{10-3x}{4})^2$

anonymous · Answer

After taking the derivative of that, I get $$A'=\frac{ 4\sqrt{3}x+9x-30 }{ 8 }$$

Am I on the right path?