I actually have two questions:

4. An open box is to be made from a rectangular piece of material 3m by 2m by cutting a congruent square from each corner and folding up the sides. What are the dimensions of the box of the largest volume made this way, and what is the volume?

5. A cylindrical container w/ a circular base is to hold 64 cubic cm. Find its dimensions so that the amt (surface area) of metal required is a minimum when the container is
a. an open cup and
b. a closed can.

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W/ 4, I have no idea how to approach. All I got is that volume of the box would be s(s-2)(s-3), I think.

As for 5, I know the formula for the volume and surface area, but my question is about the open cup for a. You probably have to subtract something from somewhere...but where? I don't really know.

Thank you very much!

4. Let each side of the square that is cut out be s m

then the box will have dimensions s by 2-2s by 3 - 2s
and the
volume = s(2-2s)(3-2s)
expand, find the derivative, set that equal to zero and solve that quadratic

This question is probably used more than any other to introduce the concept of maximum/minimum by most textbooks, only the numbers will differ.

5. another straight-forward Calculus question
let the radius be r, and the height h cm
volume = πr^2h
πr^2h = 64
h = 64/(πr^2)
a) SA = one circle + the "sleeve" of the cylinder
= πr^2 + 2πrh
= πr^2 + 2πr(64/(πr^2))
= πr^2 + 128/r

d(SA)/dr = 2πr - 128/r^2 = 0 for a max/min of SA
2πr = 128/r^2
r^3 = 64/π

take cube root, sub back into h = ....

4. To find the dimensions of the box of the largest volume, we need to find the value of 's' that maximizes the volume function.

Let's start by drawing a diagram of the rectangular piece of material with the squares cut out of each corner:

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Let the length of each side of the square cut out of each corner be 's'. Then, the dimensions of the resulting box will be:

Length = original length - 2s = 3m - 2s
Width = original width - 2s = 2m - 2s
Height = s

The volume of the box can be expressed as the product of its dimensions:

Volume = Length × Width × Height
= (3m - 2s)(2m - 2s)(s)
= 4s^3 - 10s^2 + 6s

To find the maximum volume, we can take the derivative of the volume function with respect to 's', set it equal to zero, and solve for 's':

dV/ds = 12s^2 - 20s + 6
0 = 12s^2 - 20s + 6

This equation can be factored as:

0 = 2(6s^2 - 10s + 3)
0 = 2(2s - 1)(3s - 3)

So, the possible values for 's' are:

s = 1/2 or s = 1

Since 's' represents the length of a side of the square, it cannot be negative, and hence 's = 1' is the only valid solution.

Therefore, the dimensions of the box with the largest volume are:

Length = 3m - 2s = 3m - 2(1) = 1m
Width = 2m - 2s = 2m - 2(1) = 0m
Height = s = 1m

The volume of the box is given by:

Volume = 1m × 0m × 1m = 0 cubic meters

Therefore, the dimensions of the box of the largest volume are 1m x 0m x 1m, and the volume is 0 cubic meters.

5.
a. To find the dimensions of the open cup, we need to minimize the surface area of the cylindrical container. The surface area of the cylindrical container can be calculated using the formula:

Surface Area = 2πrh + πr^2

Given that the volume of the container is 64 cubic cm, we have:

Volume = πr^2h = 64 cubic cm

We can solve this equation for 'h' in terms of 'r' as:

h = 64 / (πr^2)

Now, substitute this expression for 'h' in the surface area formula:

Surface Area = 2πr(64 / (πr^2)) + πr^2
= 128 / r + πr^2

To minimize this function, we can take the derivative of the surface area with respect to 'r', set it equal to zero, and solve for 'r':

dA/dr = -128 / r^2 + 2πr = 0
128 = 2πr^3
r^3 = 64 / π
r = (64 / π)^(1/3)

Therefore, the radius of the circular base is (64 / π)^(1/3) cm.

b. To find the dimensions of the closed can, we need to minimize the surface area of the cylindrical container including the top and bottom. The surface area of the closed can can be calculated as:

Surface Area = 2πrh + 2πr^2

Using the same volume equation as above, we can obtain the value of 'h' as:

h = 64 / (πr^2)

Substituting this expression for 'h' in the surface area equation, we have:

Surface Area = 2πr(64 / (πr^2)) + 2πr^2
= 128 / r + 2πr^2

Taking the derivative of the surface area with respect to 'r', setting it equal to zero, and solving for 'r', we get:

dA/dr = -128 / r^2 + 4πr = 0
128 = 4πr^3
r^3 = 32 / π
r = (32 / π)^(1/3)

Therefore, the radius of the circular base is (32 / π)^(1/3) cm for the closed can.

Please note that I have used the symbol 'π' to represent the mathematical constant pi.

1. Finding the dimensions and volume of the open box:

To solve this problem, we need to maximize the volume of the box. Let's assume that each side of the congruent square cut from the corners has length "s".

When we cut out the squares from each corner and fold up the sides, the dimensions of the resulting box will be:
Length = 3 - 2s
Width = 2 - 2s
Height = s

The volume of the box is given by: Volume = Length * Width * Height

Substituting the dimensions we found, the volume becomes: V = (3 - 2s)(2 - 2s)(s)

To find the maximum value for the volume, we need to find the value of "s" that maximizes the function V.

Taking the derivative of V with respect to "s" and setting it equal to zero, we get:

dV/ds = 0

Solving this equation will give us the value of "s" that maximizes the volume.

2. Finding the dimensions for the open cup and closed can:

For the cylindrical container:

a. Open cup:
To minimize the surface area of the metal required, we need to find the dimensions that minimize the surface area of the open cup. The surface area of the open cup is given by:

Surface Area = πr² + πrh

We know that the volume of the container is 64 cubic cm, given as V = πr²h.

Using this information, we can express "h" in terms of "r" from the volume equation, and substitute it into the surface area equation. This will give us a function for the surface area in terms of a single variable, "r".

To minimize the surface area, we can then take the derivative of the surface area function with respect to "r" and set it equal to zero. Solving this equation will give us the value of "r" that minimizes the surface area, and we can calculate the corresponding values of "h" and "r" for the open cup.

b. Closed can:
For a closed can, we need to consider both the surface area of the curved part (lateral surface area) and the surface area of the base.

The formula for the surface area of the cylinder is given by: S = 2πrh + πr²

Similar to the open cup, we can express "h" in terms of "r" using the volume equation V = πr²h, and substitute it into the surface area equation to get a function for the surface area in terms of a single variable, "r". By taking the derivative of this function with respect to "r" and setting it equal to zero, we can find the value of "r" that minimizes the surface area. This will give us the dimensions for the closed can.

By following these steps, we can find the dimensions for both the open cup and the closed can that minimize the surface area.