An open box is to be made out of a 10-inch by 14-inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume.
Well, well, well, it looks like we have a budding mathematician here! So, you want to find the dimensions of the box that will give you the largest volume, huh? Let's get cracking!
To begin with, let's call the length of the square that will be cut out from each corner "x". Now, after you cut out these squares, what's left is a rectangle with a length of (14 - 2x) inches and a width of (10 - 2x) inches. Are you following me so far?
Now, we need to find the volume of this open box. Remember, volume is the product of length, width, and height. In this case, the height of the box will also be the length of the squares we cut out, which is "x" inches.
So, the volume V(x) of the box is given by the formula V(x) = x * (14 - 2x) * (10 - 2x). Still with me?
Now, to find the values of x that will give us the maximum volume, we need a little bit of calculus (gasp!). We'll take the derivative of V(x) with respect to x, set it equal to zero, and solve for x. This will give us the critical points where the volume is either at a maximum or minimum.
After doing some fancy calculus, we find that x = 1 inch. Yes, just 1 tiny inch! Plug this value back into the volume equation, and voila! The resulting dimensions of the box with the largest volume are 12 inches by 8 inches by 1 inch.
So, there you have it - the clownishly calculated dimensions of the box that will maximize the volume. Happy box-making, my curious friend!
To find the dimensions of the resulting box with the largest volume, we need to follow these steps:
Step 1: Determine the dimensions
Let's assume that each side of the square cut out from each corner has a length of x inches.
Step 2: Calculate the length and width of the box
Since the original dimensions of the cardboard are 10 inches by 14 inches, and we're cutting squares of length x from each corner, the length of the resulting box will be (10 - 2x) inches, and the width will be (14 - 2x) inches.
Step 3: Calculate the height of the box
The height of the box will be x inches.
Step 4: Calculate the volume of the box
The volume of a box is given by the product of its length, width, and height. Therefore, the volume V(x) of the box is V(x) = (10 - 2x)(14 - 2x)x.
Step 5: Find the value of x that maximizes the volume
To find the value of x that maximizes the volume, we need to take the derivative of V(x) with respect to x, set it equal to zero, and solve for x.
Step 6: Calculate the dimensions of the box with the largest volume
Once we find the value of x, we can substitute it back into the expressions for the length, width, and height to get the dimensions of the box with the largest volume.
Let's go ahead and solve it step by step:
Step 1: Determine the dimensions
Let's assume that each side of the square cut out from each corner has a length of x inches.
Step 2: Calculate the length and width of the box
The length of the resulting box will be (10 - 2x) inches, and the width will be (14 - 2x) inches.
Step 3: Calculate the height of the box
The height of the box will be x inches.
Step 4: Calculate the volume of the box
The volume of a box is given by the product of its length, width, and height. Therefore, the volume V(x) of the box is V(x) = (10 - 2x)(14 - 2x)x.
Step 5: Find the value of x that maximizes the volume
To find the value of x that maximizes the volume, we need to take the derivative of V(x) with respect to x, set it equal to zero, and solve for x.
V'(x) = 0
Differentiating V(x) = (10 - 2x)(14 - 2x)x, we get:
V'(x) = [(10 - 2x)(-2) + x(-2)](14 - 2x) + (10 - 2x)(14 - 2x)(-2)
Simplifying that, we get:
V'(x) = -4(10 - 2x)(14 - 2x) + 2x(14 - 2x) - 4(10 - 2x)(14 - 2x)
= -8x(10 - 2x) + 2x(14 - 2x) - 4(10 - 2x)(14 - 2x)
= -16x^2 + 16x + 56x - 8x^2 - 4(140 - 40x - 28x + 4x^2)
= -16x^2 + 16x + 56x - 8x^2 - 560 + 160x + 112x - 16x^2
= -32x^2 + 304x - 560
Setting V'(x) = 0 and solving for x:
-32x^2 + 304x - 560 = 0
Dividing by -8:
4x^2 - 38x + 70 = 0
Factoring the equation:
(2x - 5)(2x - 14) = 0
Setting each factor to zero:
2x - 5 = 0 or 2x - 14 = 0
Solving for x:
2x = 5 or 2x = 14
x = 5/2 or x = 14/2
x = 2.5 or x = 7
However, since we're dealing with lengths, we can't have x = 2.5 because it would result in a negative length for the sides of the resulting box.
Therefore, x = 7 inches.
Step 6: Calculate the dimensions of the box with the largest volume
Substituting x = 7 back into the expressions for the length, width, and height, we get:
Length = 10 - 2x = 10 - 2(7) = 10 - 14 = -4 (not valid)
Width = 14 - 2x = 14 - 2(7) = 14 - 14 = 0 (not valid)
Height = x = 7 inches
Since the length and width cannot be negative or zero, we can conclude that the value of x = 7 inches is not valid.
Therefore, there is an error in the problem statement or our calculation.
To find the dimensions of the resulting box with the largest volume, we need to optimize the volume with respect to the dimensions of the box.
Let's assume that the side length of the square cut from each corner is "x" inches. Since we're cutting equal squares from all four corners, this will make the length of the resulting box (10 - 2x) inches and the width (14 - 2x) inches.
The height of the box is actually the side length of the squares we cut out, which is also "x" inches.
Now, we can calculate the volume of the box:
V = length × width × height
V = (10 - 2x) × (14 - 2x) × x
To find the largest volume, we need to maximize this function. We can accomplish this by taking the derivative of V with respect to x, setting it equal to zero, and solving for x.
Let's differentiate the function:
dV/dx = (14 - 2x) × (10 - 2x) + x × (-4) × (10 - 2x) + x × (14 - 2x) × (-4)
dV/dx = (14 - 2x)(10 - 2x) - 4x(10 - 2x) - 4x(14 - 2x)
dV/dx = 140 - 28x - 20x + 4x^2 - 40x + 8x^2 - 56x + 8x^2
dV/dx = 16x^2 - 144x + 140
Now, set dV/dx equal to zero and solve for x:
16x^2 - 144x + 140 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -144, and c = 140. Applying the quadratic formula:
x = (-(-144) ± √((-144)^2 - 4(16)(140))) / (2(16))
x = (144 ± √(20736 - 8960)) / 32
x = (144 ± √11776) / 32
x = (144 ± 108) / 32
This gives us two potential values for x:
x₁ = (144 + 108) / 32 = 252 / 32 = 7.875
x₂ = (144 - 108) / 32 = 36 / 32 = 1.125
Since we cannot have a negative length for the sides, x = 1.125 is not a valid solution in this context. Therefore, x = 7.875.
Now, we can substitute this value of x back into the expressions for the length, width, and height of the box:
Length = 10 - 2x = 10 - 2(7.875) = 10 - 15.75 = -5.75 (not a valid length)
Width = 14 - 2x = 14 - 2(7.875) = 14 - 15.75 = -1.75 (not a valid width)
Height = x = 7.875
Since negative lengths and widths are not valid, it means that our assumption was wrong. We made an error in assuming that cutting equal squares from all four corners would result in the maximum volume.
The correct method to find the dimensions of the resulting box with the largest volume involves considering the relationship between the dimensions and the volume geometrically:
If we draw a diagram, we can see that the length and width must be equal in order to maximize the volume. Thus, let's assume the length and width are both "l" inches. Therefore, we have:
Length = l
Width = l
Height = x
We know that the original cardboard dimensions are 10 inches by 14 inches. Therefore, the length and width of the resulting box should be less than or equal to these values:
l ≤ 10 inches
l ≤ 14 inches
Since the length and width must be equal, we can set up the inequality:
l ≤ 10 inches (since 10 ≤ 14)
Substituting the values of length, width, and height into the volume equation, we have:
V = l × l × x
V = l² × x
Now, we can optimize the volume by maximizing the function V = l² × x with respect to "l." Differentiating V with respect to l:
dV/dl = 2lx
Setting it equal to zero and solving for l:
2lx = 0
l = 0
Since l cannot be zero (as it would not form a box), this means that l must be as close as possible to zero without exceeding either the length or width of the cardboard.
Therefore, the dimensions of the resulting box with the largest volume are:
Length = Width = l (as close to zero as possible without exceeding the original dimensions of the cardboard)
Height = x (from the previous calculation, x = 7.875 inches)
L = length = 14 - 2h
w = width = 10 - 2 h
V = L * w * h
= (14-2h)(10-2h)(h)
= (140 - 28 h - 20 h + 4 h^2 )h
= 4 h^3 -48 h^2 + 140 h
so
dV/dh = 0 at max
0 = 12 h^2 - 96 h + 140
0 = 3 h^2 - 24 h + 35
h = [ 24 +/- sqrt (576-420) ] /6
h = 4 +/- (1/6)(12.5)
h = 6.08 or 1.92
6.08 is impossible so 1.92 is height