A rectangular piece of cardboard measuring 12 cm by 18 cm is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each square in centimetres.
a. Give the restrictions on x.
b. For what value of x will the volume be a maximum?
My answer:
a. 0<x<6
a. correct
b. The volume with the cuts made and the sides folded up is
v = (12-2x)(18-2x)x = 4x^3-60x^2+216x
If you have calculus, then
dv/dx = 12(x^2-10x+18)
dv/dx=0 when x=5-√7 ≈ 2.35
If no calculus, then a graphical or numeric method is needed.
To find the restrictions on x, we need to consider the dimensions of the rectangular piece of cardboard. Since we are cutting equal size squares from each corner, the length and width of the resulting box will be reduced by 2 times the side length of the square.
Therefore, for the length, we can directly subtract 2 times x from the initial length of 18 cm: 18 - 2x.
Similarly, for the width, we can subtract 2 times x from the initial width of 12 cm: 12 - 2x.
Since the resulting box cannot have negative dimensions, we must ensure that both the length and the width after the cut are positive.
So, we can set up the following inequalities:
Length after cut > 0: 18 - 2x > 0
Width after cut > 0: 12 - 2x > 0
Solving each inequality separately:
18 - 2x > 0
-2x > -18
x < 9
This gives us the restriction: x < 9.
12 - 2x > 0
-2x > -12
x < 6
This gives us another restriction: x < 6.
Combining the two restrictions, the final restrictions on x are: 0 < x < 6.
b. To find the value of x that maximizes the volume of the resulting box, we need to express the volume as a function of x and then find the maximum value of that function.
The volume of the box is given by V = length * width * height.
Since we are folding up the sides of the cardboard to create the box, the height will be equal to the side length of the squares we cut. So, the height is x cm.
The length of the resulting box is 18 - 2x cm, and the width is 12 - 2x cm.
Thus, the volume function in terms of x is: V(x) = (18 - 2x) * (12 - 2x) * x.
To find the value of x that maximizes the volume, we can take the derivative of V(x) with respect to x and set it equal to zero.
dV/dx = 0
Solve this equation to find the critical points. Then, you can evaluate the second derivative of V(x) to determine if the critical point is the maximum or minimum.
Finally, compare the values of V(x) at the critical points to find which one gives the maximum volume.
a. The restrictions on x are 0 < x < 6. This is because the side length of each square cannot be more than half the shorter side of the rectangle, which is 6 cm. Additionally, the side length of each square cannot be negative or zero.
b. To find the value of x that will maximize the volume, we first need to find the expression for the volume of the box. The volume of a rectangular box is given by V = lwh, where l, w, and h represent the length, width, and height of the box, respectively.
In this case, the length (l) of the box is 18 cm, the width (w) is 12 cm, and the height (h) is x cm (since folding up the sides of the cardboard creates the height of the box).
We want to maximize the volume (V), so let's express it in terms of x:
V = (18 - 2x)(12 - 2x)x
V = 4x(18 - 2x)(12 - 2x)
To find the value of x that will maximize the volume, we can either plot a graph of V against x and find the maximum point, or we can use calculus to find the maximum value.
Using calculus, we can find the critical points (points where the derivative is equal to zero or does not exist) and check for maximum values. The derivative of V with respect to x is:
V' = 4(18 - 2x)(12 - 2x) + 4x(-2)(12 - 2x) + 4x(18 - 2x)(-2)
V' = 4(18 - 2x)(12 - 2x) - 8x(12 - 2x) - 8x(18 - 2x)
V' = 4(18 - 2x)(12 - 2x) - 8(12x - 2x²) - 8(18x - 2x²)
V' = 4(18x - 2x²)(12 - 2x) - 96x + 16x² - 144x + 16x²
V' = -32x³ + 128x² - 192x + 864
Setting V' equal to zero and solving for x:
-32x³ + 128x² - 192x + 864 = 0
Using a graphing calculator or factoring, we can find that x = 3 is a critical point. We can also verify that 0 < x < 6 falls within the allowed range of x.
To determine if x = 3 is a maximum or minimum, we can check the second derivative. The second derivative of V with respect to x is:
V'' = -96x² + 256x - 192
Evaluating V'' at x = 3:
V''(3) = -96(3)² + 256(3) - 192
V''(3) = -864 + 768 - 192
V''(3) = -288
Since V''(3) is negative, x = 3 is a maximum point.
Therefore, the value of x that will maximize the volume of the box is 3 cm.