Question

Find the dimensions of the rectangular box of maximum volume if the sum of the length, width, and height equals 180. (Give exact answers. Use symbolic notation and fractions where needed.)

Answer 1

To find the dimensions of the rectangular box of maximum volume, we can use calculus and the method of Lagrange multipliers to solve the problem. The dimensions of the rectangular box of maximum volume are 60 cm, 60 cm, and 60 cm.

Step-by-step explanation:

To find the dimensions of the rectangular box of maximum volume, we can use calculus. Let's denote the length, width, and height as x, y, and z respectively. We are given that x + y + z = 180. The volume of the box is V = xyz. To maximize V, we need to maximize the function V = xyz subject to the constraint x + y + z = 180.

We can use the method of Lagrange multipliers to solve this problem. Let's define the Lagrangian function L = xyz + λ(x + y + z - 180), where λ is the Lagrange multiplier. Taking the partial derivatives of L with respect to x, y, z, and λ, we get:

dL/dx = yz + λ = 0

dL/dy = xz + λ = 0

dL/dz = xy + λ = 0

dL/dλ = x + y + z - 180 = 0

Solving these equations, we find that x = y = z = 60. Therefore, the dimensions of the rectangular box of maximum volume are 60 cm, 60 cm, and 60 cm.

Answer 2

Step-by-step explanation:

Let the dimension of the box be x, y and z.

Volume of the box V = xyz

Since the box is a rectangular box, the base will be a square of equal length.

V = x²h where h is the height of the box

If the sum of the length, width, and height equals 180, then x+x+h = 180

2x+h = 180

h = 180-2x

Substitute h = 180-2x into the volume of the box

V = x²(180-2x)

V = 180x²-2x³

The box has its maximum volume when
`(dV)/(dx) = 0`

`(dV)/(dx) = 360x-6x^(2)`

`360x-6x^(2) = 0\\360x = 6x^2\\360 = 6x\\6x = 360\\x = (360)/(6)\\ x = 60`

Since 2x+h = 180

Substitute x = 60 into the equation ang get the height h

2(60)+h = 180

120+h = 180

h = 180-120

h = 60

Hence the dimension of the rectangular box is 60 by 60 by 60