A rectangular box has a square base. If the sum of the height and the perimeter of the square base is 14 in, what is the maximum possible volume?
To find the maximum possible volume of the rectangular box, we need to determine the dimensions of the box first.
Let's assume that the side length of the square base is "x", and the height of the box is "h".
Given that the sum of the height and the perimeter of the square base is 14 inches, we can write the equation:
h + 4x = 14
Now, let's express the volume of the rectangular box in terms of "x" and "h".
Since the base of the box is a square and the height is "h", the volume can be calculated as V = x * x * h, or simply V = x^2 * h.
We want to find the maximum possible volume, so we need to express the volume "V" in terms of a single variable. We can rewrite the equation above as h = 14 - 4x and substitute it into the volume formula:
V = x^2 * (14 - 4x)
Now, we have a volume equation in terms of a single variable "x". To find the maximum volume, we should differentiate the volume equation with respect to "x" and set it to zero:
dV/dx = 2x(14 - 4x) + x^2(-4) = 0
Simplifying the equation:
28x - 8x^2 - 4x^2 = 0
-12x^2 + 28x = 0
4x(7 - 3x) = 0
So we have two possible solutions: x = 0 or 7/3.
Since a side length of 0 does not make sense for a square base, we discard x = 0.
Therefore, the side length of the square base is x = 7/3 inches.
Substituting this value back into the equation h + 4x = 14, we can solve for "h":
h + 4(7/3) = 14
h + 28/3 = 14
h = 14 - 28/3
h = 42/3 - 28/3
h = 14/3
So, the height of the box h is 14/3 inches.
Finally, we can find the maximum possible volume by substituting the values of "x" and "h" into the volume equation:
V = (7/3)^2 * (14/3)
Calculating further:
V = 49/9 * 14/3 = 686/27
Therefore, the maximum possible volume of the rectangular box is 25.41 cubic inches (approximately).
Let's start by assigning variables to the dimensions of the rectangular box. Let's call the side length of the square base "x" and the height of the box "h".
From the given information, we can form the equation:
x + 4x + h = 14.
Simplifying this equation, we have:
5x + h = 14.
Now, we need to express the volume of the rectangular box in terms of x and h. The volume of a rectangular box is given by the product of its dimensions, which in this case is:
Volume = x * x * h = x^2 * h.
To maximize the volume, we need to find the maximum value of x^2 * h.
Since we have the equation 5x + h = 14, we can express h in terms of x:
h = 14 - 5x.
Substituting h in terms of x into the equation for volume, we have:
Volume = x^2 * (14 - 5x).
To find the maximum volume, we need to find the value of x that maximizes this expression.
To do this, we can take the derivative of the expression with respect to x and set it equal to zero.
d(Volume)/dx = 2x(14 - 5x) + x^2(-5) = 28x - 7x^2 - 5x^2 = 28x - 12x^2.
Setting this derivative equal to zero, we have:
28x - 12x^2 = 0.
Factoring out an x, we have:
x(28 - 12x) = 0.
This equation can be solved by setting each factor equal to zero:
x = 0 or x = 28/12 = 7/3.
Since the side length of a square cannot be zero, we can disregard the x = 0 solution.
Therefore, the value of x that maximizes the volume is x = 7/3.
To find the corresponding height, we can substitute this value of x back into the equation for h:
h = 14 - 5x = 14 - 5(7/3) = 14 - 35/3 = 42/3 - 35/3 = 7/3.
Therefore, the maximum possible volume of the rectangular box is:
Volume = (7/3)^2 * (7/3) = 49/9 * 7/3 = 343/27 square inches.
h+4s = 14
v = hs^2 = (14-4s)s^2
so find s when dv/ds = 0