A SHEET OF CARDBOARD 180 INCHES SQUARE IS USED to make an open box by cutting squares of equal size from the corners and folding up the sides, what size squares should be cut to obtain a box with the largest possible volume?
If x represents the side of each of the 4 little squares cut from the corners, then the dimensions of the box would be (180-2x), (180-2x) and x.
Volume,
V(x)=x(180-2x)²
Find V'(x)=dV(x)/dx and equate to zero. Solve for x. Retain only positive root, x0.
To make sure the value of x is maximum, Find V"(x)=d²V(x)/dx² and evaluate V"(x0).
If V"(x0)<0, V(x0) is at its maximum.
Post your answer for a check if you wish.
180
To find the size of the squares that should be cut in order to obtain a box with the largest possible volume, we can follow these steps:
Step 1: Let's assume that the length of each side of the cut square is "x". This means that the length of the original square will be reduced by 2x (two sides are cut).
Step 2: The dimensions of the box will be (180 - 2x) inches by (180 - 2x) inches by x inches.
Step 3: The volume of the box can be calculated by multiplying the three dimensions:
Volume = (180 - 2x) * (180 - 2x) * x
Step 4: Now, we need to find the value of x that maximizes the volume. To do this, we can take the derivative of the volume with respect to x and set it equal to zero:
dV/dx = 0
Step 5: Taking the derivative and solving for x:
dV/dx = (180 - 2x)*(180 - 2x)*1 - 2(180 - 2x)*(180 - 2x) = 0
180^2 - 4x^2 - 4x^2 + 16x^2 = 0
180^2 - 8x^2 = 0
8x^2 = 180^2
x^2 = (180^2)/8
x^2 = 4050
x = sqrt(4050)
x ≈ 63.64
Step 6: Since the size of the squares must be a whole number, we round x to the nearest whole number:
x ≈ 64
Therefore, in order to obtain a box with the largest possible volume, squares of size 64 inches should be cut from each corner of the cardboard.
To find the size of the squares that should be cut to obtain a box with the largest possible volume, we need to first understand the dimensions and constraints of the problem.
Let's assume that the length of each side of the square to be cut is "x" inches. Based on that assumption, we can determine the dimensions of the resulting box.
If you cut a square of side "x" from each corner of the cardboard sheet, the dimensions of the resulting box will be:
Length: (180 - 2x) inches
Width: (180 - 2x) inches
Height: x inches
To find the volume of the box, we multiply the length, width, and height:
Volume = (180 - 2x) * (180 - 2x) * x
To maximize the volume, we can differentiate this equation with respect to x and find the critical points. We can then determine which critical point corresponds to the maximum volume.
Differentiating the equation:
dV/dx = 2(180 - 2x)(-2)(180 - 2x) + (180 - 2x)^2
Simplifying the equation:
dV/dx = -8(180 - 2x)^2 + (180 - 2x)^2
dV/dx = -8(180 - 2x)^2 + (180 - 2x)^2
dV/dx = (180 - 2x)^2(1 - 8)
Setting dV/dx equal to zero to find the critical points:
(180 - 2x)^2(1 - 8) = 0
Since (180 - 2x)^2 cannot be zero, we set (1 - 8) equal to zero:
1 - 8 = 0
-7 = 0
There are no solutions to the equation -7 = 0. Hence, there are no critical points in this scenario.
To determine the size of the squares that give the maximum volume, we need to check the endpoints of the interval. The interval for x in this case is [0, 90]. We substitute the endpoints into the volume equation and compare the values to find the maximum:
When x = 0:
Volume = (180 - 2 * 0)(180 - 2 * 0)(0)
Volume = 0
When x = 90:
Volume = (180 - 2 * 90)(180 - 2 * 90)(90)
Volume = 0
Both endpoints result in a volume of 0, which means that there is no maximum volume for this problem. This is likely due to the fact that using the entire cardboard sheet to make an open box does not allow for any space to enclose.
Therefore, in this particular scenario, cutting squares from the cardboard sheet will not result in a box with the largest possible volume.