## anonymous 3 years ago 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)

1. anonymous

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

2. anonymous

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

3. anonymous

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