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January 17, 2017
Posted by **Mishaka** on Friday, December 16, 2011 at 8:20pm.

So far I have: V = (14 - 2x)(8 - 2x)(h)

- Calculus (Optimization, Still Need Help) -
**Mishaka**, Friday, December 16, 2011 at 8:43pmI 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? - Calculus (Optimization) -
**bobpursley**, Friday, December 16, 2011 at 8:48pmset V'=0, and you have a quadratic. Why not use the quadratic equation..

12x^2+88x-112=0

3x^2+22x-28=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. - Calculus (Optimization) -
**Mishaka**, Friday, December 16, 2011 at 9:00pmI 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?

- Calculus (Optimization) -
**bobpursley**, Friday, December 16, 2011 at 9:04pmI don't see how you got the signs as you did. Please recheck

- Calculus (Optimization) -
**bobpursley**, Friday, December 16, 2011 at 9:06pmhang 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. - Calculus (Optimization) -
**bobpursley**, Friday, December 16, 2011 at 9:07pmOk, your equation is right. Recheck your final signs as I stated.

- Calculus (Optimization) -
**Mishaka**, Friday, December 16, 2011 at 9:12pmI rechecked and found that 3x^2-22x+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).

- Calculus (Optimization) -
**bobpursley**, Friday, December 16, 2011 at 9:17pmLet me do some thinking...

if 0=V' = 112 - 88x - 12x^2

multipy both sides by -1, and

12x^2+88x-112=0

I don't see those as your signs.... - Calculus (Optimization) -
**Mishaka**, Friday, December 16, 2011 at 9:19pmNow I'm lost, I don't get why you changed the signs.

- Calculus (Optimization) -
**bobpursley**, Friday, December 16, 2011 at 9:22pmYou 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 - Calculus (Optimization) -
**Mishaka**, Friday, December 16, 2011 at 9:28pmOkay, 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.

- Calculus (Optimization) -
**Mishaka**, Friday, December 16, 2011 at 9:31pmNevermind, 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!