Find the dimensions of the rectangular box with largest volume if the total surface area is given as 4 cm^2. (Let x,y, and z be the dimensions of the rectangular box.)
Just as max area of a rectangle occurs when it is a square, max volume occurs when the box is a cube.
So, what is the side length of a cube with area 4?
nope. There are six square faces.
To find the dimensions of the rectangular box with the largest volume, given that the total surface area is 4 cm^2, we can use a mathematical optimization technique.
Let's break down the problem into smaller steps:
Step 1: Represent the surface area equation
The total surface area of a rectangular box can be represented by the equation:
2(xy + xz + yz) = 4
Step 2: Represent the volume equation
The volume of a rectangular box can be represented by the equation:
V = xyz
Step 3: Solve for one variable in terms of the other two
From the surface area equation, let's solve for z in terms of x and y:
z = (4 - 2xy) / (2(x + y))
Step 4: Substitute the value of z in the volume equation
Substitute the value of z from step 3 into the volume equation:
V = xy((4 - 2xy) / (2(x + y)))
Step 5: Simplify the volume equation
Simplify the equation by multiplying through by (x + y):
V = (4x^2y + 4xy^2 - 2x^2y^2) / 2
Step 6: Differentiate the volume equation with respect to one variable
To find the maximum volume, we can differentiate the volume equation with respect to one variable and set it equal to zero. Let's differentiate with respect to x:
dV/dx = (8xy - 4x^2y - 4xy^2) / 2 = 4xy - 2x^2y - 2xy^2 = 0
Step 7: Solve for the critical point
Let's solve the equation 4xy - 2x^2y - 2xy^2 = 0 for the critical point (x, y). This can be factored as:
2xy(2 - x - y) = 0
So, either:
1) 2xy = 0, which means either x = 0 or y = 0
2) 2 - x - y = 0, which means x + y = 2
Step 8: Analyze the critical points
We have two cases to consider: when x or y is zero, or when x + y = 2.
Case 1: x or y = 0
If x or y is zero, then the volume V would also be zero. So, these are not the dimensions we are looking for.
Case 2: x + y = 2
If x + y = 2, it means the remaining side's length z would also be 2. So, the dimensions of the box would be x = 1, y = 1, and z = 2.
Step 9: Check the endpoints
We have also assumed that x, y, and z are positive. To confirm that this solution is the maximum volume, we need to check the endpoints as well. Since the surface area is fixed, the length of the sides cannot be negative, so the endpoints are not valid solutions in this case.
Therefore, the dimensions of the rectangular box with the largest volume, given the total surface area of 4 cm^2, are x = 1 cm, y = 1 cm, and z = 2 cm.