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I need some help optimizing the volume of a box. I've followed Optimization, volume of a box this question along while doing my problem and felt confident I was getting the correct answer. But still, I submitted it in my homework and got it wrong. I double checked, followed it again and can't find where I'm going wrong.

So I have 900cm^2 to use to maximize the volume of a box with no top. My constraint is

$$b^2 + 4bh = 900$$

I solved for H in the constraint and plugged it into the volume equation, simplified to give me.

$$V(b) = \frac{1}{4}b(900b - b^3)$$

I found the critical points of the derivative, giving me [0, 17.32, 300]. When plugged into V(b), 0 and 300 both give smaller values than 17.32. This makes it a maximum and thus is the spot where the volume is correctly optimized, right?

So then I plug 17.32 into my H function

$$h = (900 - b^2) / 4b$$

This gives me 8 as the height of my sides, and my base is 17.32. Plugging these into the volume function gives me

$$V = (17.32)^2 \cdot 8 = 2399.86 \, \text{cm}^2$$

Yet, this is wrong. Can anyone highlight just where I'm going off here?

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Paul
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  • The value you get for $h$ is not exactly 8. It's $\frac{150}{\sqrt{300}}=\frac{15}{\sqrt{3}} \approx 8.66$ – JSQuareD Apr 08 '16 at 18:55
  • Wow! I feel extremely stupid. Overlooked this multiple times, thank you so much! – Paul Apr 08 '16 at 18:56
  • Also, in your formula for $V(b)$, you have an extra factor $b$. But it seems that was just a typo. – JSQuareD Apr 08 '16 at 18:57
  • The box in the linked question had a square base. Is yours constrained to have a square base? You don't state this in your question. – John Apr 08 '16 at 19:03
  • Yes, it does have a square base. After fixing my value for the sides from 8 to 8.66, I had the correct answer. I simply overlooked the decimals for whatever reason. – Paul Apr 08 '16 at 19:04

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