Question

A piece of wire 10m 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 are enclosed is (a) a maximum? (B) Is a minimum?
Help please

Answer 1

@amistre64

Answer 2

The total volume it going to be"\[Volue_{total} = Volume_{\square} +Volume_{\triangle}\]

Answer 3

the constraints on these volume is the perimeter. Set the sides of one of the object to a variable, say, x. then the perimeter will be related through the length of original wire:

Let x = length of side on square
Perimeter of square: = 4x
Perimeter of Tri = 10m - 4x
Side of Triangle = (10-4x)/3

Answer 4

Now express the Volume (the quantity you wish to maximize) as a function of the variable:
\[Volume_{total} = volume_{\square}+Volume_{\triangle}\]
\[Volume_{total} =(length*width)_{\square}+\frac{1}{2}(base*height)_{\triangle}\]
\[Volume_{total} = (x*x)+\frac{1}{2}(\frac{10-4x}{2})(\sqrt3(\frac{10-4x}{2}))\]

Answer 5

Why are you setting it equal to the volume?

Answer 6

Because the question asks for you to solve for the maximum/minimum area enclosed depending on how you cut the wire. To solve Max/min question you need an equation that you can take the derivative of, in this case Volume with respect to length

Answer 7

derivatives allow you to solve max/min because when the derivative is zero, that represents a hump (max/min) in the original function

Answer 8

thanks a lot, I get it now

Answer 9

hahah..i see. I've written volume when I've meant Area throughout the whole question. please substitute Area

Answer 10

One more question, how did you get the height of the triangle?

Answer 11

Ya, I figured!!

Answer 12

special triangle (30 degrees/60 degrees)
or straight pythagorean relation