Question

a piece of wire 60 inches long is cut into five sections, two of length a and three of length b. The two sections of length a are bent into the form of a equilateral triangle, and the two triangles are then joined by the 3 remaining sections of length b to make a frame for a model of a triangle prism. What is the maximum volume of the triangular prism?

Answer 1

would the volume=(1/2)(sqrt(3))r^2b

Answer 2

https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=volume+of+a+triangular+prism

Answer 3

60=6a+3b

Answer 4

a=10-b/2

Answer 5

hold on

Answer 6

okay

Answer 7

This is what it looks like. I had to draw it in my head

Answer 8

yes that is how it looks like on my paper

Answer 9

do we have #'s?

Answer 10

60in wire cut into 5 sections, two of the length a and three of length b

Answer 11

I am pretty sure we can solve algebraically. you could also use the area of a right triangle, multiply by 2 and integrate over the z axis.

Answer 12

couldn't I do optimization?

Answer 13

Maybe. Its been awhile since I've done optimization. https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=volume+of+a+triangular+prism

Answer 14

use the formula in this image

Answer 15

okay

Answer 16

I don't think that formula gives me the correct answer.

Answer 17

\[2A +3B = 60\]
Let x be length of side of triangle, where A = 3x
\[6x +3B = 60\]
solve for B
\[B = 20 -2 x = 2(10-x)\]
Set up volume equation (area triangle*lengthB)
\[V = \frac{\sqrt{3}}{4}x^2B\]
\[V = \frac{\sqrt{3}}{2}x^2 (10-x)\]
Now maximize volume with respect to x
Take derivative and set equal to 0
\[\frac{dV}{dx} = \sqrt{3}x(10-x) - \frac{\sqrt{3}}{2} x^2 = 0\]
\[\rightarrow x = \frac{20}{3}\]
Finally, plug in x value into volume equation to obtain max Volume