'Simple' Optimization Problem

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The discussion revolves around solving an optimization problem involving a piece of cardboard measuring 14 inches by 10 inches, where corners are cut to form an open box. The volume equation derived is V = (10-2x)(14-2x)x, leading to a cubic function for volume. The critical points calculated are x = 6.08 and x = 1.92, with 6.08 deemed impossible due to physical constraints, leaving 1.92 as the feasible height for maximum volume. The user expresses discomfort with the numbers obtained, but finds reassurance through comparing volumes at heights slightly above and below the calculated value. Ultimately, fatigue from studying is acknowledged as a factor in the user's uncertainty.
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Just reviewing for a chapter test... I've always found optimization problems easy, but I don't have answers for these review questions so I thought I'd check my work on here.

Homework Statement


A piece of cardboard is 14 inches by 10 inches and you are going to cut out the corners and fold up the sides to form an open box. Determine the height of the box that will give a maximum volume.

Homework Equations



V = lwh

The Attempt at a Solution



Using
(10-2x) as l
(14-2x) as w
and x as the height:

V = (10-2x)(14-2x)x = 4x^3 - 48^2 +140x

dv/dx = 12x^2 - 96x + 140

0 = 12x^2 - 96x +140

x = 6.08, 1.92 or 1/3(12±√39)

... this is where I get lost. I feel like I messed it up somewhere because I get weird numbers for x. Technically, the 6.08 number is impossible, so x would have to equal 1.92. But still, I feel uncomfortable with these numbers.
 
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What makes you uncomfortable about the answer you got?

Would it make you more comfortable to compare the volume if the height were slightly above or below the height you calculated to the volume you got?
 
jamesrc said:
Would it make you more comfortable to compare the volume if the height were slightly above or below the height you calculated to the volume you got?

haha, yes, actually. It helped quite a bit. I think I'm just too tired from too many late nights of studying... thanks for the help.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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