Solve Lowest Cost Cuboid Volume Problem - Philipp

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Homework Help Overview

The problem involves optimizing the cost of a cuboid with a square base and an open top, given a fixed volume of 2 cubic meters. The cost of the material for the sides is twice that of the base material.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive equations for volume and surface area to minimize cost. Some participants suggest rearranging equations to simplify the problem and substituting variables. Others raise questions about finding critical points and handling derivatives.

Discussion Status

The discussion includes attempts to clarify the approach to the problem, with some guidance on rearranging equations and taking derivatives. There is ongoing exploration of methods to find minimum values, but no consensus has been reached on the best approach.

Contextual Notes

Participants are working within the constraints of the problem's requirements, including the fixed volume and differing material costs. There is mention of confusion regarding the handling of derivatives and critical points.

philipp2020
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hi

today i have another problem where i am not sure

Here is the question:

A cuboid, made of an square ground and which is open on the top should have the volume of 2 squaremeters. The costs for the material of the 4 sides have the double price as for the material of the square ground. Search for the cuboid with the lowest cost...

So I formed 2 equations for this problem:

1: V = x^2 * h -----> 2 = x^2 * h --->

2: 2 x^2 = 4 * x* h ------> h = 1/2 x

At the end I received a result for a length of 2 meter for square side's bottom and a height of 0.5 meter. The result seems to be right. But I don't know if the way I solved it afterwards is right or not.

Can somebody show me the right forms for the way to the result?

Thanks very much

Philipp
 
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Your first equation is good, but you can rearrange it to solve for h. This will allow you to make a substitution in your second equation so you'll only be dealing with one variable.

For your second equation, you need the surface area, which is:

[tex]x^2 + 4xh[/tex]

Since the side material is twice as expensive as the bottom material, the cost of the surface area material will be the cost of the bottom material (unimportant) times this:

[tex]x^2 + 2 * 4xh[/tex]

Substitute for h (from your first equation).

Take the derivative.

Set the derivative equal to zero.

You will wind up with the same answer you gave, although you didn't really explain how you got there.
 
and then out of x^2 + 16 *1/x

derivate:

2x - 16x^-2 = 0

An then i try to find the 0 points of this equation with a discriminant?

But how can I if there is a minus exponation. How can I find the minimum now?
 
ok thanks i found the way. Just multiply with x^2 and there it is

aaaaaaaaaaaaaaaaaaaaaaaaaaaa
 

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