Calculus (Optimization)
posted by Mishaka .
A rectangular piece of cardboard, 8 inches by 14 inches, is used to make an open top box by cutting out a small square from each corner and bending up the sides. What size square should be cut from each corner for the box to have the maximum volume?
So far I have: V = (14  2x)(8  2x)(h)

I just wanted to correct something for my equation, it should be: V = (14  2x)(8  3x)(x), which simplifies to V = 112x  44x^2  4x^3.
Take the derivative:
V' = 112  88x  12x^2
Now all I need are the roots, any help? I think I found one around 1.10594, possibly? 
set V'=0, and you have a quadratic. Why not use the quadratic equation..
12x^2+88x112=0
3x^2+22x28=0
x= (22+sqrt (22^2+12*28))/6
doing it in my head, I get about..
(22+28)/6= 4/6, 7.5 in my head.
check my work and estimates. 
I think that you might have gotten the equation wrong, I think that it should be: 3x^2  22x + 28. When I put this equation into the quadratic equation, I got 5.694254177 and 1.639079157. So the squares that need to be cut out should have an area of approximately 2.69 square inches?

I don't see how you got the signs as you did. Please recheck

hang on, I reread the problem statement. In the first response I gave, I took your equation. I don't think it is right.
give me a minute. 
Ok, your equation is right. Recheck your final signs as I stated.

I rechecked and found that 3x^222x+28 has the correct signs. Knowing this equation and the values I found from the quadratic equation, would you say that the 1.639079157 term is correct? (The 2.69 square inches came from squaring the 1.639079157).

Let me do some thinking...
if 0=V' = 112  88x  12x^2
multipy both sides by 1, and
12x^2+88x112=0
I don't see those as your signs.... 
Now I'm lost, I don't get why you changed the signs.

You are lost. This is algebra.
if 0=112  88x  12x^2
do whatever you know to put it in standard form, ax^2+bx+c=0
when you do that a and b will have the SAME signs. Surely you can do that.
if a=12, then b=88, and c=112
if a=12, then b=88, and c=112 
Okay, so does this change my original answer of approximately 1.64 to 4.42??? The 4.42 came from putting your new values in the quadratic equation.

Nevermind, that 4.42 was a mistake and my very original answer of 1.105940354 was absolutely correct!!! This is the right answer, I know it!
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