a rectangular piece of cardboard measuring 14 inches by 27 inches is to be made into a box with an open top by cutting squares of equal size from each corner and golding up the sides . let x represent the length of a side of each square. for what value of x will the volume be a maximum? round to two decimal places.

v = x(14-2x)(27-2x)

dv/dx = 2(6x^2 - 82x + 189)

dv/dx=0 when x = (41 ± √547)/6 = 2.93, 10.73

knowing the shape of cubics, you should have no trouble "differentiating" the max from the min.

length = z = 27 -2x

width = y = 14 - 2x

v = (27-2x)(14-2x)x
v = ( 378 - 82 x + 4 x^2 ) x
v = 378 x -82 x^2 + 4 x^3

max when dv/dx = 0
dv/dx = 378 - 164 x + 12 x^2 = 0
6 x^2 - 82 x + 189 = 0

x = [ 82 +/- sqrt (6724-4536) ]/12

x = [82 +/- 46.8 ]/12

x = 10.7 too big, negative width
x = 2.93 inches

To find the value of x that will maximize the volume of the box, we need to follow these steps:

Step 1: Determine the dimensions of the box after cutting squares from each corner.
- The length of the box is given as 27 inches.
- The width of the box is given as 14 inches.
- By cutting squares of equal size from each corner, the length and width of the box will be reduced by 2 times the value of x (one from each corner).

So, the length of the box = 27 - 2x.
And the width of the box = 14 - 2x.

Step 2: Calculate the height of the box.
- The height of the box will be the value of x.

Step 3: Write down the formula for the volume of the box.
- The volume of a rectangular prism is given by V = length * width * height.

So, the volume, V = (27 - 2x) * (14 - 2x) * x.

Step 4: Simplify the volume formula.
- V = (27 - 2x)(14 - 2x)x
- Expand the expression: V = 378x - 54x^2 - 28x^2 + 4x^3
- Collect like terms: V = 4x^3 - 82x^2 + 378x

Step 5: Find the maximum volume by taking the derivative.
- Differentiate the volume function with respect to x: V' = 12x^2 - 164x + 378
- Set V' equal to zero and solve for x to find critical points.

12x^2 - 164x + 378 = 0

Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a

x = (164 ± √((-164)^2 - 4 * 12 * 378)) / (2 * 12)
x = (164 ± √(26896 - 18144)) / 24
x = (164 ± √(8752)) / 24
x ≈ 6.37 or x ≈ 5.96

Step 6: Determine the value of x that will maximize the volume.
- Since x represents the length of the sides of the squares, it cannot be negative.
- The maximum value of x is 6.37, but it is not possible because it exceeds half of the length of the cardboard (14 inches). So, we discard this value.
- Thus, the value of x that will maximize the volume is approximately 5.96 inches.

Therefore, the value of x that will maximize the volume is approximately 5.96 inches.

To find the value of x that maximizes the volume, we need to follow these steps:

Step 1: Visualize the problem.
- Imagine the rectangular piece of cardboard with dimensions 14 inches by 27 inches.
- Cut squares with side length x from each corner.
- Fold up the remaining sides to form an open-top box.

Step 2: Determine expressions for the box dimensions.
- The length and width of the resulting box will be:
Length: (original length) - 2x
Width: (original width) - 2x
Height: x

Step 3: Write the volume formula.
- The volume of the box is given by V = Length × Width × Height.
- Substitute the expressions from Step 2 into the volume formula:
V = (14 - 2x)(27 - 2x)x

Step 4: Simplify the volume formula.
- Multiply the binomials:
V = (378 - 54x - 28x + 4x^2)x
V = (4x^3 - 82x^2 + 378x)

Step 5: Find the derivative of the volume formula.
- Calculate the derivative of V with respect to x.
V'(x) = 12x^2 - 164x + 378

Step 6: Set the derivative equal to zero and solve for x.
- To find the critical points where the slope changes, set V'(x) = 0:
12x^2 - 164x + 378 = 0

Step 7: Solve for x using the quadratic formula.
- Use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 12, b = -164, and c = 378.
x = (-(-164) ± √((-164)^2 - 4(12)(378))) / (2(12))
x = (164 ± √(26896 - 18144)) / 24

Step 8: Simplify the value of x.
- x = (164 ± √(8748)) / 24
- x ≈ (164 ± 93.48) / 24

Step 9: Evaluate the two possible values of x.
- x1 = (164 + 93.48) / 24 ≈ 10.11
- x2 = (164 - 93.48) / 24 ≈ 2.86

Step 10: Determine the maximum volume.
- Calculate the volume for both x1 and x2.
V1 ≈ (4(10.11)^3 - 82(10.11)^2 + 378(10.11)) ≈ 2927.47
V2 ≈ (4(2.86)^3 - 82(2.86)^2 + 378(2.86)) ≈ 1787.17

Step 11: Compare the volumes and select the maximum.
- Since we rounded to two decimal places, the maximum volume occurs when x ≈ 10.11 inches, with a volume of approximately 2927.47 cubic inches.

Therefore, the value of x that maximizes the volume is approximately 10.11 inches.