- #1
ardentmed
- 158
- 0
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.
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.