sandy is making a closed rectangular jewwellery box with a square base from two different woods . the wood for top and bottom costs $20/m^2. the wood for the side costs $30/m^2 . find dimensions that minimize cost of wood for a volume 4000cm^3?

lele

To find the dimensions that minimize the cost of wood for a volume of 4000 cm^3, we need to first determine the dimensions of the rectangular jewelry box.

Let's assume the length of the base is L, and the width of the base is also L since it is a square base. The height of the box will be H.

The volume of a rectangular box can be calculated as follows: V = L * L * H

Given that the volume is 4000 cm^3, we have the equation: 4000 = L^2 * H

Next, we need to find the cost of the wood.

The cost for the top and bottom of the jewelry box (which is the square base) is $20/m^2, while the cost for the sides is $30/m^2.

The total cost of the wood can be calculated as follows:

Cost = (2 * (L * L) * 20) + (4 * (L * H) * 30)

Simplifying this equation, we get:

Cost = 40L^2 + 120LH

Now, we have two equations:

4000 = L^2 * H (from the given volume)

Cost = 40L^2 + 120LH

To minimize the cost, we can solve these equations for L and H.

Let's start by solving the volume equation for H:

H = 4000 / (L^2)

Substitute this value of H in the cost equation:

Cost = 40L^2 + 120L * (4000 / L^2)
Cost = 40L^2 + 4800000 / L

To minimize the cost, we can differentiate the cost equation with respect to L and set the derivative equal to zero:

d(Cost)/dL = (80L - 4800000 / L^2) = 0

Simplifying further, we get:

80L - 4800000 / L^2 = 0

Multiplying through by L^2:

80L^3 - 4800000 = 0

Solving this equation for L, we find:

L^3 = 60000

Taking the cube root of both sides:

L = ∛(60000)
L ≈ 38.73 cm

Now, substitute this value of L into the volume equation to find H:

H = 4000 / (L^2)
H = 4000 / (38.73^2)
H ≈ 3.37 cm

Therefore, the dimensions that minimize the cost of wood for a volume of 4000 cm^3 are approximately:

Length (L) ≈ 38.73 cm
Width (L) ≈ 38.73 cm
Height (H) ≈ 3.37 cm

To minimize the cost of wood, we need to find the dimensions of the jewelry box that result in the least amount of wood being used.

Let's assume that the dimensions of the square base are x by x (since it's a square) and the height is h.

To find the volume of the jewelry box, we'll use the formula for the volume of a rectangular prism: V = length \times width \times height. In this case, the length (x), width (x), and height (h) are all equal.

Given that the volume of the jewelry box is 4000 cm^3, we have the equation:

4000 = x \times x \times h

Simplifying, we get:

4000 = x^2 \times h

Next, to find the cost of the wood, we need to determine the surface area of each side of the jewelry box.

The top and bottom have the same dimensions, so each has an area of x \times x = x^2.

There are four sides to the jewelry box, so we have a total of four side surfaces, each with an area of x \times h.

The total surface area of the jewelry box is:

2(x^2) + 4(xh) = 2x^2 + 4xh

Given that the wood for the top and bottom costs $20/m^2, and the wood for the sides costs $30/m^2, we can calculate the cost of the wood as follows:

Cost = (2x^2 + 4xh) \times ($20/m^2) + (2x^2) \times ($30/m^2)

To minimize the cost, we need to differentiate the cost equation with respect to one of the variables (x or h) and set the derivative equal to zero, then solve for the variable.

Differentiating the cost equation with respect to x gives:

dCost/dx = 4x + 4h = 0

Differentiating the cost equation with respect to h gives:

dCost/dh = 4x = 0

Setting both derivatives equal to zero, we have:

4x + 4h = 0 (Equation 1)
4x = 0 (Equation 2)

From Equation 2, we know that x = 0. However, since we're dealing with the dimensions of a physical object, x cannot be zero. Therefore, we can ignore Equation 2.

Solving Equation 1 for h, we get:

4h = -4x
h = -x

Now, substituting this value of h into the volume equation:

4000 = x^2 \times h
4000 = x^2 \times (-x)
4000 = -x^3

Simplifying, we find:

x^3 = -4000

Taking the cube root of both sides, we get:

x = -16.85

Since we're working with physical dimensions, the length cannot be negative. Therefore, we disregard the negative value.

Thus, the only possible solution is x ≈ 16.85.

Now, substituting this value of x back into the equation for h:

h = -x
h ≈ -16.85

Again, since the dimensions must be positive, we disregard the negative value.

Therefore, the dimensions that minimize the cost of the wood for a volume of 4000 cm^3 are approximately x = 16.85 cm and h = 16.85 cm.

let the base be x m by x m

let the height be y m

1 m^3 = 100^3 cm^3 = 1 000 000 cm^3
V= 4000 cm^3 = .004 m^3

V = x^2y
.004 = x^2y
y = .004/x^2

cost = 20(2x^2) + 30(4xy)
= 40x^2 + 120x(.004/x^2)
= 40x^2 + .048/x
d(cost)/dx = 80x - .048/x^2 = 0 for max/min of cost
80x = .048/x^2
x^3 = .0006
x = .0843 m , x = 8.43 cm
y = .5623 m , y = 56.23 cm

check my arithmetic, looks like a weird shape for a jewelry box.

check:
if x= .0843, y = .5623, then V = .004
cost = 40(.0843)^2 + .048/.0843 = .8536

try an x value slightly higher
let x = .085
cost = 40(.085)^2 + .048/.085 = .8537 , which is higher
try an x value slightly lower
let x = .084
cost = 40(.084)^2 + .048/.084 = .85366 which is also higher