# wire problem max and min

A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

(a) How much wire should be used for the square in order to maximize the total area?

I thought the answer was: 29cm due having to use the whole wire to maximize the total area but I was wrong. Can someone help me out please.

(b How much wire should be used for the square in order to minimize the total area?

I said 10 because I found x=2.50 so 2.50(4)=10 but again I was mistaken.

• What do you know of calculus? – user228113 Nov 20 '16 at 1:13
• If you want to max/min a function, you should start by writing it as a function of some variable. For instance, if $x$ is the amount of wire used for the square, then the square will have a side-length of $x/4$ and an area of $x^2/16$. That leaves $29-x$ of wire for the perimeter of the triangle, and its area can be readily computed from that. Combining these gives you your objective function $A(x)$ to be min-maxed. – Semiclassical Nov 20 '16 at 1:37
• It's not an answer. It's a question: these exercises all boil down to writing the formula for the thing you want and look for the sign of the derivative. Of course, if you do not know that, one needs to come up with a trick. Describing effectively your attempts of solution is useful for others to understand how much you know. – user228113 Nov 20 '16 at 1:41

Approach this by saying that length $x$ is used to make the square. It must be divided into four sides, so length of a side is ${x} \over 4$ and area is ${x^2} \over 16$.
The remainder length $(29-x)$ is used to make a triangle. Each side is ${29-x} \over 3$. Area of triangle is given by $\frac 12 ab \sin C = \frac 12 \frac{(29-x)^2} 9 \sin 60^o = \frac{(29-x)^2\sqrt3} {18}$
Total area is thus $\frac {x^2} {16}+\frac{(29-x)^2\sqrt3} {18}$