Question

Optimization problem: I did it, just need someone to see what I did is correct. :p

Answer 1

Uploading a photo of my answers...

Answer 2

derivative steps...

Answer 3

\[V = x(24-2x)^2\]
\[\frac{dV}{dx}=2(24-2x)(-2)(x) +(24-2x)^2\]
...
\[\frac{dV}{dx}=12x^2-192x+576\]

Answer 4

my derivation step includes product rule [ (uv)'=u'v+uv' ] but best practice would be to expand expression to individual terms and take the derivative of that instead

Answer 5

So, when I write V in terms of x,
\[x(24-2x)^2\]
I don't distribute the x
\[(24x-2x^2)^2\]
Then, take the derivative of that? I should take the derivative of when I first write V in terms of x?

Answer 6

So, I get
(x-12)(x-4)
x=4,12: critical numbers
\[y''=24x-192\]
y''=24(12)-192>0 Minimum So, it doesn't work, right?
y''=24(4)-192<0 Maximum
Then, V is maximum when x=4 @1024in^2

Answer 7

looks good.
right, initially you brought the x into the brackets but treated it as if there were no power on the brackets. later when you took the derivative with chain rule, you forgot to multiply by the derivative of what was in the brackets

if x were 12 there would be a nothing left of length and width so the Volume would be zero, a minimum.