Max volume problem, intro calc

In summary, the problem asks for the dimensions of a rectangular box with the greatest volume that can be made from 100 sq inches of cardboard, with a base that is twice as long as it is wide. The equation for volume is V = lwh and the equation for area is A = lw. After drawing a diagram and setting the base length as 2n and width as n, the volume is calculated to be 2n^2*h. However, since the box has no top, the area of the material is equal to the surface area of the box.
  • #1
katia11
18
0

Homework Statement


What are the dimensions of the base of the rectangular box of the greatest volume that can be constructed from 100 sq inches of cardboard if the base is to be twice as long as it is wide? Assume the box has no top.


Homework Equations


V box = lwh
A = lw ?


The Attempt at a Solution


I did all the other similar problems assigned but am not sure where to begin here.

Drew a diagram. Base length= 2n width=n
volume=lwh
volume=2n*n*h
volume=2n^2*h

but now what. . .am I all wrong?

Sorry if this is way easy, it's just been awhile since I did these.
 
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  • #2
Since all your material is going into making the box, the area of your material is the surface area of your box. Note that it doesn't have a top.
 

Related to Max volume problem, intro calc

1. What is the "Max Volume Problem"?

The "Max Volume Problem" is a mathematical problem in introductory calculus that involves finding the maximum volume of a three-dimensional shape, such as a cylinder or cone, given certain constraints.

2. How is the "Max Volume Problem" solved?

The "Max Volume Problem" is typically solved using the techniques of calculus, specifically optimization. This involves setting up an equation for the volume of the shape, taking the derivative, and solving for the critical points where the derivative is equal to zero. The maximum volume can then be found by plugging these critical points into the original volume equation.

3. What are some common real-world applications of the "Max Volume Problem"?

The "Max Volume Problem" has many real-world applications, such as determining the maximum amount of water that can be held in a cylindrical tank, finding the most efficient shape for a shipping container, or optimizing the size of a storage container for a given amount of material.

4. Can the "Max Volume Problem" be solved using other methods besides calculus?

While calculus is the most common and efficient way to solve the "Max Volume Problem", there are other methods that can be used, such as algebraic manipulation or geometric reasoning. However, these methods may be more time-consuming and may not always yield the most accurate solution.

5. Are there any limitations or assumptions to consider when solving the "Max Volume Problem"?

When solving the "Max Volume Problem", it is important to consider any limitations or assumptions, such as the shape being symmetrical, the material being uniform, and the shape being able to hold a certain amount of volume without collapsing. These factors can affect the accuracy of the solution and should be taken into account when solving the problem.

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