What is the optimal volume of a box with given dimensions using calculus?

In summary, the conversation is about a math problem involving finding the length of corner pieces cut out of a rectangular box to maximize its volume. The initial attempt at a solution is to set up an equation for the volume and differentiate it, but there are errors in the equation and clarification is needed on certain variables.
  • #1
Elihu5991
33
0

Homework Statement


SEE QUESTION IMAGE


Homework Equations


SEE ABOVE


The Attempt at a Solution


SEE WORKINGS IMAGE
 

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  • #2
Can you type in your work?

ehild
 
  • #3
Sorry that my handwritiing is too messy and scan isn't done right. Just trying to get this done ASAP.

(25-2x)(40-2x)

y=(25-2x)(-2)+(40-2x)(-2)
=-50+4x-40+2x
=-90+6x
x=15

Not sure what to next do.
 
  • #4
Elihu5991 said:
Sorry that my handwritiing is too messy and scan isn't done right. Just trying to get this done ASAP.

(25-2x)(40-2x)

y=(25-2x)(-2)+(40-2x)(-2)
=-50+4x-40+2x
=-90+6x
x=15

Not sure what to next do.

Is x the length of the corner pieces cut out? If so, you need to define it in words; a formulation without an explanation is worth 0.

Also, is (25-2x)*(40-2x) the thing you want to maximize? Why? What does the box look like if you use your proposed solution of x = 15?
 
  • #5
Elihu5991 said:
Sorry that my handwritiing is too messy and scan isn't done right. Just trying to get this done ASAP.

(25-2x)(40-2x)

y=(25-2x)(-2)+(40-2x)(-2)
You mean y', not y.

Also, I'm guessing that you want to maximize the volume. If so, then
y = (25-2x)(40-2x)
as the volume equation would be wrong. You're missing the height. What would it be?
 

Related to What is the optimal volume of a box with given dimensions using calculus?

1. What is optimisation using calculus?

Optimisation using calculus is a mathematical method used to find the maximum or minimum values of a function. It involves using derivatives to determine the slope of a curve and identify points where the slope is equal to zero, which correspond to maximum or minimum values.

2. Why is optimisation using calculus important?

Optimisation using calculus is important because it allows us to solve real-world problems that involve finding the best possible outcome. This can be applied in various fields such as economics, engineering, and science to make informed decisions and improve processes.

3. What are the key steps in optimisation using calculus?

The key steps in optimisation using calculus include identifying the objective function, finding the first and second derivatives of the function, setting the first derivative equal to zero to find critical points, evaluating the second derivative at these points to determine whether they are maximum or minimum values, and finally, checking the endpoints of the given interval to ensure the global maximum or minimum is found.

4. Can optimisation using calculus be applied to multivariable functions?

Yes, optimisation using calculus can be applied to multivariable functions. In this case, the partial derivatives of the function are used to find critical points, and the Hessian matrix is used to determine whether these points are maximum or minimum values.

5. Are there any limitations to optimisation using calculus?

One limitation of optimisation using calculus is that it requires the function to be continuous and differentiable in the given interval. This means that it may not be applicable to some real-world problems that involve non-differentiable functions or discontinuities. Additionally, it may not always provide the most optimal solution, as there could be multiple critical points or the global maximum/minimum may not exist.

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