a closed box has a fixed surface area A and a suqare base wide side X
a) find the formula for its volume, v, as a function x
b) sketch the graph of v against x
c) find the max value of V
Let the height be y
then A = 2x^2 + 4xy
4xy = A - 2x^2
y = (A - 2x^2)/(4x), remember A is a constant.
a) Volume = x^2y
= x^2(A - 2x^2)/(4x)
= (Ax - 2x^3)/4
b) we can't do sketching on here
c) V = (Ax - 2x^3)/4
dV/dx = A/4 - (3/2)x^2 = 0 for max of V
(3/2)x^2 = A/4
x^2 = A/6
x = √(A/6) or √A/√6
V = [(A(√A/√6) - 2(A/6)(√A/√6)]/4
= A√A/(6√6)
Check my algebra, I should have written it down on paper.
If in your answer key you don't see a radical in the bottom, they might have rationalized the denominator.
To find the formula for the volume of the box as a function of x, we need to make use of the given information. Let's break it down step by step:
a) Formula for the Volume, V, as a function of x:
The box has a square base, which means all sides of the base are of length x. The remaining sides of the box are vertical and have the same width as the base. Let's denote the height of the box as h.
Since the box has a fixed surface area A, we can break down the surface area into its respective components. The surface area of the base is x * x = x^2, and the area of the four vertical sides (all of the same width) will be 4 * x * h = 4xh.
The fixed surface area A is then the sum of these two components:
A = x^2 + 4xh
To find the volume, V, we multiply the base area (x^2) by the height (h):
V = x^2 * h
Now we need to express h in terms of x. From the surface area equation, we have:
A = x^2 + 4xh
Rearranging the equation, we can solve for h:
4xh = A - x^2
h = (A - x^2) / (4x)
Substituting this expression for h into the formula for volume, we get:
V = x^2 * [(A - x^2) / (4x)]
Simplifying further, we can cancel out the x in the numerator and denominator:
V = (x^2 * (A - x^2)) / 4
So, the formula for the volume, V, as a function of x is:
V = (x^2 * (A - x^2)) / 4
b) Sketching the graph of V against x:
To sketch the graph, we plot the volume V on the y-axis and the base width x on the x-axis. The shape of the graph will depend on the values of A.
c) Finding the maximum value of V:
To find the maximum value of V, we can differentiate the formula for V with respect to x, set the derivative equal to zero, and solve for x. The value of x for which this occurs will give us the maximum volume.
Differentiating V with respect to x:
dV/dx = (2x(A - x^2) - 4x^3) / 4
dV/dx = (2Ax - 2x^3 - 4x^3) / 4
dV/dx = (6x^2 - 2Ax) / 4
Setting the derivative equal to zero:
(6x^2 - 2Ax) / 4 = 0
Simplifying the equation:
6x^2 - 2Ax = 0
2x(3x - A) = 0
This leads to two potential solutions:
1) x = 0
2) 3x - A = 0
We can disregard x = 0 as it would result in a volume of 0.
Solving 3x - A = 0:
3x = A
x = A/3
So, the maximum value of V occurs when x = A/3.
Note: The value of A determines the behavior of the graph and the maximum volume.