Calculus help (Optimization Integration)

[Ipquarx]

I started off with

w*3w*(4w-60)=0

If anything it would simplify to minimizing 3w^2*(60-4w), not 3w^2*(4w-60). I think that's why you got a negative number.
l+w+h=60
h=60-l-w
h=60-4w

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[Pastrey Crust]

ah yes
silly me

I feel stupid for screwing up in silly things.

[Texan101]

Yes, that.
Rectangular prism. L = 3w L+W+H = 60

I got a negative answer for some reason. I started off with

w*3w*(4w-60)=0

With the equation w*3w*(4w-60), we can simplify to easily find w. 3w^2 * (4w-60) is equivalent, and gives us three answers. Either 3w^2 = 0 (The first two of our answers, since this could normally be positive or negative), or 4w = 60, which comes to w = 15.

I highly doubt this was your mistake.

[Ipquarx]

With the equation w*3w*(4w-60), we can simplify to easily find w. 3w^2 * (4w-60) is equivalent, and gives us three answers. Either 3w^2 = 0 (The first two of our answers, since this could normally be positive or negative), or 4w = 60, which comes to w = 15.

I highly doubt this was your mistake.

No, we're trying to maximize 3w²(60-4w), not make it 0.

[Texan101]

No, we're trying to maximize 3w²(4w-60), not make it 0.

Alright, that's a better challenge.
So, distributing 3w^2 yeilds 12w^3-60w^2
The derivative, in relation to w, is 36w^2-120w

There is no maximum to this equation. Only a minimum.

[Ipquarx]

There is no maximum to this equation. Only a minimum.

Remember, we're doing this within a specific domain here: (0, 60]
So since it just goes up, the answer is w=60, or 1,944,000.

Note that you may only be able to get rid of 1 variable, and this would be a 2 variable optimization problem, but i'm not certain.

[Texan101]

Remember, we're doing this within a specific domain here: (0, 60]
So since it just goes up, the answer is w=60, or 1,944,000.

Note that you may only be able to get rid of 1 variable, and this would be a 2 variable optimization problem, but i'm not certain.

Sorry. Didn't see the restricted domain.