Derivative Applications Question (High-school Calculus)

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Homework Help Overview

The problem involves finding the dimensions of an open-top box with a square base that minimizes surface area while maintaining a fixed volume of 4500 cm³. The subject area is calculus, specifically the application of derivatives in optimization problems.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the relationships between the variables for volume and surface area. There are attempts to derive the surface area equation and its derivative. Some participants question the validity of multiplying by x² in the context of the surface area equation.

Discussion Status

The discussion has progressed with participants providing feedback on each other's attempts. There is a recognition of improved approaches to finding the derivative, and some participants express agreement with the proposed critical number. However, there is no explicit consensus on the final dimensions or solution.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an ongoing discussion about the correctness of mathematical manipulations and assumptions made during the problem-solving process.

NDiggity
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Homework Statement


A box with a square base and an open top has a volume of 4500cm^3. What are the dimensions of the box that will minimize the amount of material used? (Remember that the amount of material used refers to the surface area of the box).

My buddy and I spent a solid hour on this questions with no luck. Please help!

Homework Equations


No set equations, make your own.

The Attempt at a Solution



First we set our variables:
Let x=a side of the base
Let y=height of the box

Then we isolated the y variable:
x*x*y=4500
x^2*y=4500
y=4500/x^2

Then we made a surface area equation:
S(x)=x^2 + 4xy
S(x)=x^2 + 4x(4500/x^2)
S(x)=x^2 + 18000x/x^2
Then we multiplied everything by x^2 to get rid of the denominator to get:
S(x)=x^4 + 18000x
S(x)=x(x^3 + 18000)

Then we found the derivative of this equation:
S'(x)=4x^3 + 18000
S'(x)=4(x^3 + 4500)

This is where we got stuck because the critical number will be negative. Please help, we do not know what we did wrong.
 
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Then we multiplied everything by x^2 to get rid of the denominator to get:
If you're going to do that you need to multiply S(x) as well as you are not multiplying by one.
 
NDiggity said:
Then we isolated the y variable:
x*x*y=4500
x^2*y=4500
y=4500/x^2

Then we made a surface area equation:
S(x)=x^2 + 4xy
S(x)=x^2 + 4x(4500/x^2)
S(x)=x^2 + 18000x/x^2
So far, so good. Simplify this and find its minimum.

Then we multiplied everything by x^2 to get rid of the denominator to get:
S(x)=x^4 + 18000x
S(x)=x(x^3 + 18000)
Say what now? You can't just arbitrarily multiply by x^2! For one thing, x^2*S(x) does not equal S(x).
 
Ok so if I don't multiply by x^2 and simplify, I'm left with:
S(x)=x^2 + 18000/x
S(x)=x^2 + 18000x^-1

So with the derivative I get:
S'(x)=2x -18000x^-2
S'(x)=2x -18000/x^2

Is this better?
 
Much better. :smile:
 
Yay, thank you very much for the help so far! So, would my critical number be the cube root of 9000? If so my dimensions are cube root of 9000 or 20.8 cm by 10.4cm.
 
Last edited:
Looks good to me. :smile:
 
yeah correct
 

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