MATH SOLVE

10 months ago

Q:
# The material for the base of an open box costs 1.5 times what the material for its side faces costs. Find the dimensions of the box with the maximum volume that can be built with a fixed cost C. Solve it by the grange multiplier or by the criterion of second partial derivatives.

Accepted Solution

A:

Let the length, width, and height of the box be x, y, and z, respectively. Let C be the fixed cost. The volume of the box is given by:
V = xyz
The cost of the material for the base is 1.5xy, and the cost of the material for the side faces is 2yz+2xz. Therefore, the total cost is given by:
C = 1.5xy + 2yz + 2xz
We want to maximize V subject to the constraint C. We can use the Lagrange multiplier method to solve this problem. The Lagrangian is given by:
L(x, y, z, \lambda) = xyz + \lambda(C - 1.5xy - 2yz - 2xz)
Taking partial derivatives with respect to x, y, z, and λ, we get:
L_x = yz - 1.5xy \lambda = 0
L_y = xz - 1.5xy \lambda = 0
L_z = xy - 2yz \lambda - 2xz \lambda = 0
L_\lambda = C - 1.5xy - 2yz - 2xz = 0
Solving the above equations for x, y, z, and λ, we get:
x = \frac{C}{6 \lambda}
y = \frac{C}{4 \lambda}
z = \frac{C}{3 \lambda}
\lambda = \frac{1}{C}
Substituting these values into the equation for V, we get:
V = \frac{C^3}{216}
To maximize V, we want to maximize C
3
. Since C is fixed, the maximum value of V is achieved when C is as large as possible. Therefore, the optimal dimensions of the box are:
x = y = z = \frac{C}{6}
Solution by the criterion of second partial derivatives:
Let the length, width, and height of the box be x, y, and z, respectively. Let V be the volume of the box. The volume is given by:
V = xyz
We want to maximize V subject to the constraint that the cost of the material is fixed. The cost of the material is given by:
C = 1.5xy + 2yz + 2xz
We can use the criterion of second partial derivatives to solve this problem. The Hessian of V is given by:
H = \begin{pmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{pmatrix}
The determinant of the Hessian is equal to xyz
2
. Since z is always positive, the Hessian is positive definite if and only if xy>0. Therefore, the maximum value of V is achieved when xy>0.
In order to satisfy the constraint on the cost of the material, we must have 1.5xy+2yz+2xz=C. We can substitute the equation z=C/(3xy) into this equation to get:
1.5xy + 2y(C / (3xy)) + 2x(C / (3xy)) = C
Simplifying this equation, we get:
xy = C / 24
Since we want xy>0, we must have x>0 and y>0. Therefore, the optimal dimensions of the box are:
x = y = z = \frac{C}{6}
Both solutions give the same answer, so we can conclude that the dimensions of the box with the maximum volume that can be built with a fixed cost C are x=y=z= c/6