A cardboard box without a top is to have volume 500000 cubic cm. Find the dimensions which minimize the amount of material used. List them in ascending order.

To minimize the amount of material used, we need to find the dimensions of the cardboard box that result in the smallest surface area.

Let's say the dimensions of the box are length (L), width (W), and height (H).

Given that the volume is 500000 cubic cm, we have the equation:

L * W * H = 500000

We want to minimize the surface area, which is a combination of the bottom area, top area, and the four side areas.

The surface area equation is given by:

Surface Area = 2(LW + LH + WH)

We need to express one of the variables in terms of the other two, so we can substitute it into the surface area equation. Let's solve the volume equation for H:

H = 500000 / (LW)

Substitute this value of H into the surface area equation:

Surface Area = 2(LW + L(500000 / LW) + W(500000 / LW))

Simplifying the equation further:

Surface Area = 2(LW + 1000000/L + 1000000/W)

To minimize the surface area, we need to minimize this equation. To find the minimum, we can take the derivative of the equation with respect to L and W. Then set the derivatives equal to zero and solve for L and W.

Differentiating the surface area equation with respect to L gives:

d(Surface Area) / dL = 2(W - 1000000/L^2) = 0

Simplifying the equation:

W - 1000000/L^2 = 0
W = 1000000/L^2

Differentiating the surface area equation with respect to W gives:

d(Surface Area) / dW = 2(L - 1000000/W^2) = 0

Simplifying the equation:

L - 1000000/W^2 = 0
L = 1000000/W^2

Now we have two equations:

W = 1000000/L^2
L = 1000000/W^2

Let's substitute L from the first equation into the second equation:

1000000/W^2 = 1000000/(1000000/W^2)^2

Simplifying this equation gives:

1000000/W^2 = 1000000 * W^4 / 1000000^2

Simplifying further:

W^6 = 1000000^2

Taking the sixth root of both sides:

W = (1000000^2)^(1/6) = 1000000^(2/6) = 1000000^(1/3) = 100

Substituting this value of W into the equation W = 1000000/L^2:

100 = 1000000/L^2

Simplifying this equation gives:

L^2 = 1000000/100 = 10000

Taking the square root of both sides:

L = 100

Now we have the dimensions of the cardboard box:

L = 100 cm
W = 100 cm
H = 500000 / (LW) = 500000 / (100 * 100) = 50 cm

Therefore, the dimensions of the cardboard box that minimize the amount of material used are 100 cm x 100 cm x 50 cm, listed in ascending order.

To minimize the amount of material used, we need to minimize the surface area of the cardboard box.

Let's denote the dimensions of the base of the box as follows:
- Length: L cm
- Width: W cm

Since the box doesn't have a top, we need to consider that only the bottom and sides will be made of cardboard.

The surface area of the bottom is L * W square cm.

The surface area of the sides consists of two rectangles with dimensions:
- Length: L cm, Height: H cm
- Width: W cm, Height: H cm

So, the total surface area of the cardboard box is given by the sum of the surface area of the bottom and the surface area of the two sides:

Surface Area = (L * W) + 2(L * H) + 2(W * H)

Given that the volume of the box is 500000 cubic cm, we have:

Volume = L * W * H = 500000 cubic cm

To find the dimensions that minimize the material used, we need to minimize the surface area. However, we have two unknowns (L and W) in our equation.

We can use the relationship between volume and dimensions to express one variable in terms of the other. Let's solve the volume equation for H:

H = 500000 / (L * W)

Substitute this value of H back into the surface area equation to have a single-variable equation:

Surface Area(L, W) = (L * W) + 2(L * (500000 / (L * W))) + 2(W * (500000 / (L * W)))

Simplifying this equation gives us:

Surface Area(L, W) = L * W + 1000000 / L + 1000000 / W

To minimize the surface area, we need to find the critical points by taking the partial derivatives with respect to L and W, and setting them equal to zero:

d(Surface Area) / dl = W - (1000000 / L^2) = 0
d(Surface Area) / dw = L - (1000000 / W^2) = 0

Solving these two equations simultaneously will give us the values of L and W that minimize the surface area.

Let's solve the first equation: W - (1000000 / L^2) = 0

W = 1000000 / L^2

Substitute the value of W into the second equation: L - (1000000 / (1000000 / L^2)^2) = 0

L - (1000000 / (1000000 / L^4)) = 0

L - L^4 = 0

Factoring out L from the equation gives us: L(1 - L^3) = 0

So, either L = 0 or 1 - L^3 = 0

Since L cannot be zero in this context, we need to solve 1 - L^3 = 0.

1 - L^3 = 0

L^3 = 1

Taking the cube root of both sides, we get: L = 1

Now, substitute this value of L into the equation we obtained for W: W = 1000000 / L^2

W = 1000000 / 1^2

W = 1000000

So, the dimensions that minimize the amount of material used are L = 1 cm and W = 1000000 cm, listed in ascending order.

Note: It is worth noting that the value of W is disproportionately large due to the absence of a top on the cardboard box, resulting in a base that is almost infinitely long in one dimension.

A.

You can assume symmetry between length and width, which reduces to the width(=length) and the height.
Furthermore, one of the two can be eliminated from the volume relation:
w^2h=500000
So the minimization problem is reduced to one single dimension as in elementary calculus.

B.
The same results can be obtained by calculating the area of material required:
A=2h(b+w)+bw + L(bwh-500000)
the second term introduces the Lagrange multiplier.
Take partial derivatives with respect to w,b,h and L and solve for each variable from the 4 equations.
This method should give the same results as in part A.