AN OPEN BOX IS TO BE MADE FROM A CARDBOARD 20 INCHES BY 14 INCHES, BY CUTTING A SQUARE FROM EACH CORNER AND FOLDING UP THE SIDES. FIND THE DIMENSIONS OF THE BOX THAT WILL MAXIMIZE THE VOLUME OF THE BOX
To find the dimensions of the box that will maximize the volume, we need to follow these steps:
Step 1: Let's assume that the length of the square to be cut from each corner is 'x' inches. Therefore, the dimensions of the resulting box will be:
Length = 20 - 2x inches
Width = 14 - 2x inches
Height = x inches
Step 2: Now, we can calculate the volume of the box using the formula:
Volume = Length * Width * Height
Volume = (20 - 2x) * (14 - 2x) * x
Step 3: To maximize the volume, we can take the derivative of the volume function and set it equal to zero. So, let's differentiate the volume function with respect to 'x':
dV/dx = (14 - 2x)(20 - 2x) + x(-4)(20 - 2x) = 0
Step 4: Simplify the equation:
(14 - 2x)(20 - 2x) - 4x(20 - 2x) = 0
(14 - 2x)(20 - 2x) - 80x + 8x^2 = 0
Step 5: Expand and simplify further:
280 - 40x - 28x + 4x^2 - 80x + 8x^2 = 0
12x^2 - 148x + 280 = 0
Step 6: Solve the quadratic equation:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-(-148) ± √((-148)^2 - 4*12*280)) / (2*12)
x = (148 ± √(21904 - 13440)) / 24
x = (148 ± √8464) / 24
x = (148 ± 92) / 24
So, we have two possible values for 'x':
x1 = (148 + 92) / 24 = 9.17 inches (rounded to two decimal places)
x2 = (148 - 92) / 24 = 2.5 inches (rounded to one decimal place)
Step 7: Since the dimensions of the cardboard are limited to 20 inches by 14 inches, we can only consider the value of 'x' that is within this range. Therefore, x = 2.5 inches.
Step 8: Now that we have the value of 'x', we can calculate the dimensions of the box:
Length = 20 - 2*2.5 = 20 - 5 = 15 inches
Width = 14 - 2*2.5 = 14 - 5 = 9 inches
Height = 2.5 inches
Therefore, the dimensions of the box that will maximize the volume are 15 inches by 9 inches by 2.5 inches.
To find the dimensions of the box that will maximize its volume, we need to take the following steps:
Step 1: Define the variables:
Let's denote the length of the square that will be cut from each corner of the cardboard as "x". This means that the length of the side of the box will be "20 - 2x" and the width will be "14 - 2x".
Step 2: Determine the volume of the box:
The volume of the box can be calculated by multiplying its length, width, and height. Since the height is the same as the length of the square that will be cut from the corners, it will also be "x". Therefore, the volume of the box is given by:
Volume = (20 - 2x)(14 - 2x)(x)
Step 3: Expand and simplify the volume function:
Multiply the terms using the distributive property to expand the volume function:
Volume = (20 - 2x)(14 - 2x)(x)
= (280 - 28x - 40x + 4x^2)(x)
= (280 - 68x + 4x^2)(x)
= 4x^3 - 68x^2 + 280x
Step 4: Maximize the volume:
To find the maximum value, we need to find the critical points. Take the derivative of the volume function and set it equal to zero:
Volume' = 12x^2 - 136x + 280
12x^2 - 136x + 280 = 0
Step 5: Solve for x:
To solve the quadratic equation, we can use the quadratic formula:
x = (-(-136) ± sqrt((-136)^2 - 4(12)(280))) / (2(12))
x = (136 ± sqrt(18496 - 13440)) / 24
x = (136 ± sqrt(50456)) / 24
Calculating the values using a calculator, we find two possible values for x: approximately 1.8134 and 9.1866.
Step 6: Evaluate the critical points:
To find the maximum volume, we need to evaluate the volume function at both critical points (x = 1.8134 and x = 9.1866). The highest volume will give us the optimal dimensions of the box.
Plug in the value of x back into the volume function:
For x = 1.8134:
Volume = 4(1.8134)^3 - 68(1.8134)^2 + 280(1.8134) ≈ 183.52
For x = 9.1866:
Volume = 4(9.1866)^3 - 68(9.1866)^2 + 280(9.1866) ≈ 186.11
Therefore, the dimension of the box that will maximize the volume is approximately:
Length = 20 - 2x ≈ 20 - 2(1.8134) ≈ 16.37 inches
Width = 14 - 2x ≈ 14 - 2(1.8134) ≈ 10.37 inches
Height = x ≈ 1.8134 inches
Hence, the dimensions of the box that will maximize the volume are approximately 16.37 inches by 10.37 inches by 1.8134 inches.
OK! STOP SHOUTING
If an x-inch square is cut from each corner, the volume is
v = (20-2x)(14-2x)*x = 4(x^3 - 17x^2 + 70x
dv/dx = 4(3x^2 - 34x + 70)
So max volume is when x = (17-√79)/3
now you can finish it off