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..
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^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).
Let me do some thinking...
if 0=V' = 112 - 88x - 12x^2
multipy both sides by -1, and
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!