Consider a rectangle cardboard box without top and bottom. The diagonal of the box has length 1. Use lagrange multipliers to find the maximum surface area of the paper used to make this box. What are the dimensions of this optimal box?
To find the maximum surface area of the paper used to make the box, we need to set up the problem using Lagrange multipliers.
Let's start by defining the dimensions of the box. Let's call the length of the box L, the width of the box W, and the height of the box H.
Since the box is a rectangular cardboard box without top and bottom, the surface area of the box can be calculated by summing the areas of the four sides of the box.
The surface area of the box is given by:
S = 2(LW + LH + WH)
However, we also have the constraint that the diagonal of the box has a length of 1. Using the Pythagorean theorem, the diagonal can be expressed as:
L^2 + W^2 + H^2 = 1
Now, let's set up the Lagrangian:
Lagrange function (F) = S + λ(D - 1)
where λ (lambda) is the Lagrange multiplier and D is the constraint equation.
To find the maximum surface area, we need to find the critical points where the gradient of the Lagrange function is zero.
∇F = ∇S + λ∇D = 0
Taking the partial derivatives with respect to L, W, H, and λ, we have:
∂F/∂L = 2W + λ(2L) = 0 (1)
∂F/∂W = 2L + λ(2W) = 0 (2)
∂F/∂H = 2LH + λ(2H) = 0 (3)
∂F/∂λ = D - 1 = 0 (4)
From equations (1) and (2), we can solve for λ:
2W + 2λL = 0
2L + 2λW = 0
Dividing both equations by 2, we get:
W + λL = 0
L + λW = 0
Solving these equations simultaneously, we eliminate λ:
W^2 = L^2
W = ±L
Using equation (4), we can substitute W = ±L:
L^2 + L^2 + H^2 = 1
2L^2 + H^2 = 1
Now let's solve for H:
H^2 = 1 - 2L^2
H = ±√(1 - 2L^2)
So, we have two possible solutions:
Solution 1: L, W, H = L, -L, ±√(1 - 2L^2)
Solution 2: L, W, H = -L, L, ±√(1 - 2L^2)
To determine the maximum surface area, we need to evaluate the Lagrange function at these critical points and compare the values.
Plug the values of L, W, and H into the surface area equation:
S = 2(LW + LH + WH)
For Solution 1:
S1 = 2(L)(-L) + 2(L)(±√(1 - 2L^2)) + (-L)(±√(1 - 2L^2))
S1 = -2L^2 + 2L√(1 - 2L^2) ± √(1 - 2L^2)
For Solution 2:
S2 = 2(-L)(L) + 2(-L)(±√(1 - 2L^2)) + (L)(±√(1 - 2L^2))
S2 = -2L^2 - 2L√(1 - 2L^2) ± √(1 - 2L^2)
Take the derivative of S1 and S2 with respect to L, and set them equal to zero to find the maximum surface area.
dS1/dL = -4L + 2√(1 - 2L^2) - (4L^2)/(√(1 - 2L^2))
dS2/dL = -4L - 2√(1 - 2L^2) + (4L^2)/(√(1 - 2L^2))
Solving these equations for L will give us the dimensions of the optimal box.