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bazinga276
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)
put the side of the triangle as \(x\) and the side of the square is \(y\) then \[3x+4y=10\] solve for \(y\) get \[y=\frac{10-3x}{4}\]
total area is therefore what you said triangle area is \(A(x)=\frac{\sqrt{3}x^2}{4}+(\frac{10-3x}{4})^2\)
After taking the derivative of that, I get \[A'=\frac{ 4\sqrt{3}x+9x-30 }{ 8 }\] Am I on the right path?