If 1,200 cm^2 of material is available to make a box with a square base and an open top, find the maximum volume of the box in cubic centimeters. Answer to the nearest cubic centimeter.
You want a square base box with a fixed surface area:
SA = s² + 4sh
1200 = s² + 4sh
What's the maximum volume you can make with that box?
V = s²h
Solve the first equation for h, then substitute and solve for s into the volume equation:
1200 = s² + 4sh
-4sh = s² - 1200
h = -s/4 + 300/s
V = s²h
V = s²(-s/4 + 300/s)
V = -s³/4 + 300s
Now that we have volume in terms of one variable, we can find its maximum by taking the first derivative of that function, set it to zero, then solve for s:
dV/ds = -(3/4)s² + 300
0 = -(3/4)s² + 300
(3/4)s² = 300
s² = 400
s = 20
(throwing out the negative value of "s" since we're dealing with a box)
Now that we have "s", substitute it back into the volume equation to determine what that volume is:
V = -s³/4 + 300s
V = -(20)³/4 + 300(20)
V = -8000/4 + 6000
V = -2000 + 6000
V = 4000
The maximum volume is 4000 cm³ when the side of the box is 20cm² and the height is 10 cm.
V = s²h
4000 = 20²h
4000 = 400h
h = 10
To find the maximum volume of the box, we need to optimize the dimensions of the box. Let's begin by determining the side length of the square base.
Let's call the side length of the square base "x" centimeters. Since the box has a square base, its surface area, excluding the top, is given by 4x^2 (four sides of the square base).
The remaining material will be used to form the sides and bottom of the box. Therefore, the height of the box will be (1200 - 4x^2)/x.
The volume of the box is given by the product of the base area and the height: V = x^2*(1200 - 4x^2)/x = x(1200 - 4x^2).
To maximize the volume, we need to find the value of x that maximizes the expression x(1200 - 4x^2). This can be done by taking the derivative of the expression with respect to x and finding the critical points.
Let's calculate the derivative: dV/dx = 1200 - 12x^2.
Setting the derivative equal to zero: 1200 - 12x^2 = 0.
Rearranging the equation: 12x^2 = 1200.
Dividing both sides by 12: x^2 = 100.
Taking the square root of both sides: x = ±10.
Since the side length cannot be negative, we discard x = -10.
Therefore, the critical point is x = 10 centimeters.
To confirm that this is a maximum and not a minimum, we can take the second derivative of the expression: d²V/dx² = -24x.
Substituting x = 10 into the second derivative: d²V/dx² = -24(10) = -240.
Since the second derivative is negative, it confirms that x = 10 gives us the maximum volume.
Now, we can substitute x = 10 into the expression for volume to find the maximum volume: V = 10(1200 - 4(10)^2) = 10(1200 - 400) = 10(800) = 8000 cubic centimeters.
Therefore, the maximum volume of the box is 8000 cubic centimeters.