# Optimization of a piece of wire used to construct a square and a rectangle, minimize area

A piece of wire is to be used to construct a square and a rectangle. The width of the rectangle will be 1/3 of the length of the rectangle. The length of the wire is 20 inches. Find the dimensions of the square and the rectangle if we want to minimize area.

• Welcome to math.SE. What have you tried – Shailesh Apr 3 '17 at 3:20
• Read the problem. Let $w$ be the width of the rectangle. What is the area of the rectangle? How much wire is left? What is the area of the square? Add that to the area of the rectangle, differentiate, set to zero... – Ross Millikan Apr 3 '17 at 3:22
• I have tried solving it a few times but my answers didn't make sense. The constraint equation I set up is 20= 4x+2L+2(1/3L) and the optimization equation A=x^2+L(1/3L). Where X is a side of the square and L is the length of the rectangle. I solved the constraint for X and then plugged it into the optimization equation, then took the derivative and set that equal to zero to solve for L. I am not quite sure what I am doing wrong. – user431805 Apr 3 '17 at 3:28
• Well if $H=L/3$ why not have $L=3H$ So $20=4x+6H+2H= 4(x+2H)$. And $A=x^2+4H$. Now try and solve $A$ with respect to $x$ given these two equations. – Sentinel135 Apr 3 '17 at 3:32

The perimeter of both the square and rectangle is, $4x+2H+2H=20$. We're trying to minimize area so $s^2+HL=A$ and $\min\{A\}$. Let $H=\frac{L}{3}$, then $L=3H$. Thus by substitution we have $$4x+8H=20\rightarrow x+2H=5\\ s^2+4H=A$$ This implies that $H= \frac{5-x}{2}$. So through substitution we have $$x^2-2x+10=A$$ From here you can differentiate the equation, set it to zero, and then find your solutions easily.