MHB Minimizing Material Used for a Box

[ardentmed]

Hey guys,

I'm having trouble with this problem set I'm working on at the moment. I'd appreciate some help with this question:

(I'm only asking about number two, not one. Thanks.)

The lack of a top is accounted for in the formula for the box's surface area, which should be:

SA= L^2 +4HL
And the volume is simply:
32,000 = HL^2

THerefore, solving for the derivative of SA (or dSA) should give:

3L^2 - L^3 - 12,800.
L = 18.56635.

Ergo,
h=92.831.

Therefore, the dimension should be:
18.6cm x 92.8cm x 18.6cm

Am I close? Are the significant figures off?
Thanks in advance.

 

[Dethrone]

The length you got is wrong. Can you show me the your steps?

$$SA (l) = l^2 + \frac{4(32000)}{l}$$

Take its derivative and tell me what you get.

 

[ardentmed]

The length you got is wrong. Can you show me the your steps?

$$SA (l) = l^2 + \frac{4(32000)}{l}$$

Take its derivative and tell me what you get.

I believe that the derivative is:

dSA = 2l - 128,000/(l^2)

Multiply the expression by the common denominator and solve for l.

Am I on the right track? What do I do from here?

 

[Dethrone]

Yes, that is right. What do you get for the length?