Question

Calculus optimization problem

Answer 1

A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximum? b) a minimum?

Answer 2

Here is my work so far:
Let x be the side of the square and y the side of the triangle.
4x + 3y = 10
y = (10 - 4x)/3
\[A = x^2 + \frac{ \sqrt3 }{ 4 }y^2\]
Substitute y.
\[A = x^2+ \frac{ \sqrt3 }{ 4 }(\frac{ 10-4x }{3 })^2\]
Then I have to take the derivative right?

Answer 3

here's a more intuitive way to look at the problem. We seek to minimize the perimeter area ratio of a given segment of wire with no bounds. A square's perimeter area ratio is
p:(p/4)^2 or 1:p/16 or 16:p.
A circle's perimeter to area ratio is p:(p/(2Pi))^2*Pi or 1:p/(4Pi) which is 4Pi:p

Now a circle is really a regular polygon as the number of sides approaches infinity, so, you'd expect the regular polygon with more sides to enclose more area. And if so, why bother wasting wire on another polygon?

But yes, you have to take the derivative if you want to do it the sad calculus way

Answer 4

Would this be the derivative?
\[dA/dt = 2x + -2\sqrt3 (\frac{ 10-4x }{ 3 })\]

Answer 5

@zepdrix I could REALLY use some help on this... :(

Answer 6

Sorry I need to start from the beginning just so I can follow along a little better :U

Answer 7

So here is our cut.
Making a cut of length x, and then the remaining portion.