Problem solving with derivatives.
A rectangular box has square base of edge length x cm. Its framework of 12 edges is constructed from wire of total length 36cm. Find:
i. the height of the box in terms of x ?
ii. the volume of the box in terms of x?
ii. the value for x which the volume is a maximum.
top 4 x
bottom 4 x
corners vertical 4 h
8 x + 4 h = 36
h = (36 - 8 x)/4 = 9 - 2 x (part i.)
v = x^2 h = x^2 (9 - 2 x) = 9 x^2 - 2 x^3
dv/dx = 0 for max or min
dv/dx = 18 x- 6 x^2 = 0
6 x ( 3 - x) = 0
x = 0 is minimum of course
x = 3 is maximum then
i. To find the height of the box in terms of x, we need to consider the length of wire used to construct the framework. The framework consists of 12 edges, so we have a total of 12x cm of wire.
The wire is used to form the base of the box and the four vertical edges. Since the base is a square with an edge length of x cm, we can use 4x cm of wire for the vertical edges.
Therefore, the remaining wire, 12x - 4x = 8x cm, is used to form the four vertical edges. Since there are four equal vertical edges, each one has a length of 2x cm.
The height of the box is thus equal to the length of one of these vertical edges, which is 2x cm.
ii. The volume of a rectangular box is given by the formula V = lwh, where l, w, and h are the length, width, and height, respectively. In this case, since the base is a square, the length and width are both equal to x cm.
Therefore, the volume V of the box is V = (x)(x)(2x) = 2x^3 cm^3.
iii. To find the value for x at which the volume is a maximum, we can take the derivative of the volume function and set it equal to zero.
dV/dx = 6x^2
Setting this derivative equal to zero, we have:
6x^2 = 0
Solving for x, we find that x = 0.
However, this result does not make sense in this context because we are dealing with physical dimensions. Since we cannot have a box with zero edge length, there is no real solution for x at which the volume is a maximum.
So, it seems we don't have a clear maximum for the volume of the box in terms of x. Perhaps it's time to think outside the box and consider other factors or constraints in the problem!
To solve this problem, we can start by understanding the given information and forming equations.
Let's denote the height of the box as h cm.
i. To find the height of the box in terms of x, we need to consider the framework of the box, which consists of 12 edges. Each edge forms one side of the square base and one side of the rectangular height.
Since the square base has edge length x cm, the perimeter of the square base is 4x cm. This means that 4x cm of wire is used for the base edges.
Since we have a total of 12 edges in the framework, the remaining wire length is given by (36 - 4x) cm.
This remaining wire length is used to form the four vertical edges of the rectangular height. Since there are four equal vertical edges, each has length (36 - 4x)/4 = (9 - x) cm.
Since the height of the box is the length of one of these vertical edges, we can say that the height h in terms of x is h = 9 - x cm.
ii. To find the volume of the box in terms of x, we need to multiply the base area by the height. The base area is given by (x * x) = x^2 cm^2. Therefore, the volume V in terms of x is V = x^2 * (9 - x) cm^3.
iii. To find the value for x at which the volume is maximum, we can find the critical points by taking the derivative of the volume equation with respect to x, and then setting it equal to zero.
Taking the derivative of V = x^2 * (9 - x) with respect to x, we get:
dV/dx = 2x(9 - x) + x^2(-1)
= 18x - 2x^2 - x^2
= 18x - 3x^2
Now, setting dV/dx = 0:
18x - 3x^2 = 0
3x(6 - x) = 0
This equation can be satisfied when either 3x = 0 or 6 - x = 0.
From 3x = 0, we find that x = 0.
From 6 - x = 0, we find that x = 6.
To determine which value of x gives the maximum volume, we can check the second derivative of the volume equation.
Taking the second derivative of V = x^2 * (9 - x) with respect to x, we get:
d^2V/dx^2 = 18 - 6x
Now, substituting x = 6 into d^2V/dx^2:
d^2V/dx^2 = 18 - 6(6)
= 18 - 36
= -18
Since the second derivative is negative, it indicates that x = 6 gives a point of maximum for the volume of the box.
Therefore, the value for x at which the volume is maximum is x = 6 cm.
To solve this problem, we'll use the concept of derivatives to find the height, volume, and the value of x that maximizes the volume of the box.
Before we start, let's work out the equations and variables involved:
Let x be the length of the edge of the square base of the box (in cm).
Let h be the height of the box (in cm).
Let V be the volume of the box (in cubic cm).
Now let's solve each part of the problem step by step:
i. Finding the height of the box in terms of x:
In a rectangular box, the height can be determined by subtracting twice the length of the base from the total length. Since there are 12 edges, and each has a length of 36 cm, the total length of wire is 12 times the length of each edge:
12x = 36
To find the height, we subtract twice the length of the base (2x) from the total length:
h = 36 - 2x
So, the height of the box in terms of x is given by h = 36 -2x.
ii. Finding the volume of the box in terms of x:
The volume of a rectangular box is the product of its length, width, and height. Since the base is square, the length and the width are the same and given by the side length x. Therefore, the volume is given by:
V = x * x * h
V = x^2 * h
Substituting the expression for the height obtained in part i, we have:
V = x^2 * (36 - 2x)
So, the volume of the box in terms of x is given by V = x^2 * (36 - 2x).
iii. Finding the value for x at which the volume is maximum:
To find the value of x that maximizes the volume, we need to find the critical points of the volume function and determine which one corresponds to a maximum.
To find the critical points, we take the derivative of the volume function with respect to x, and then set it equal to zero:
dV/dx = 2x(36 - 2x) + x^2(-2)
= 72x - 4x^2 - 2x^2
= -6x^2 + 72x
Now, setting the derivative equal to zero:
-6x^2 + 72x = 0
Factoring out -6x:
-6x(x - 12) = 0
Setting each factor equal to zero, we find two critical points:
-6x = 0, which gives x = 0
x - 12 = 0, which gives x = 12
Now, we need to determine which value of x corresponds to a maximum volume. To do this, we can use the second derivative test. We take the second derivative of the volume function and evaluate it at our critical points:
d^2V/dx^2 = -12x + 72
For x = 0, the second derivative is positive (d^2V/dx^2 = 72), so this critical point corresponds to a local minimum rather than a maximum. Therefore, we can discard x = 0 as a solution.
For x = 12, the second derivative is negative (d^2V/dx^2 = -12), which indicates a maximum.
Therefore, the value of x which maximizes the volume is x = 12.
By plugging this value of x back into the volume equation we obtained in part ii (V = x^2 * (36 - 2x)), we can calculate the maximum volume of the box.