Take an 8.5 by 14-inch piece of paper and cut out four equal squares from the corners. Fold up the sides to create an open box. Find the dimensions of the box that has maximum volume. (Enter your answers as a comma-separated list. Round your answer to three decimal places.)
cut squares of length x
v = x(8.5-2x)^2 = 72.25x - 34x^2 + 4x^3
dv/dx = 72.25 -68x + 12x^2
max/min achieved at x = 17/12 or 17/4
Just knowing about the shape of cubics, it's clear that 17/12 is where the max volume occurs.
So, the box is 8.5-17/6 x 8.5-17/6 x 17/12 = 17/3 x 17/3 x 17/12
To find the dimensions of the box with maximum volume, we need to set up an equation for the volume of the box and then find the critical points by taking the derivative.
Let's start by figuring out the dimensions of the base of the box. Since we're cutting equal squares from the corners, the length of each side of the base after folding up will be (8.5 - 2x), and the width will be (14 - 2x), where x is the length of the side of the square cut from the corners.
The height of the box will be equal to the length of the side of the square cut from the corners, which is x.
So, the volume of the box can be expressed as:
V = (8.5 - 2x)(14 - 2x)x
To find the dimensions of the box that maximize the volume, we need to maximize the volume function V with respect to x. To do this, we'll find the critical points by taking the derivative of V with respect to x, setting it equal to zero, and solving for x.
Let's find the derivative:
dV/dx = (14 - 2x)(8.5 - 2x) + x(-4)(14 - 2x) + (8.5 - 2x)(-4)
Simplifying the expression:
dV/dx = (14 - 2x)(8.5 - 2x) - 4x(14 - 2x) - 4(8.5 - 2x)
Now, set the derivative equal to zero and solve for x:
(14 - 2x)(8.5 - 2x) - 4x(14 - 2x) - 4(8.5 - 2x) = 0
Expanding and simplifying the equation:
(2x^2 - 18x + 59.5) - (28x - 4x^2) - (34 - 8.5x) = 0
Rearranging the terms:
-2x^2 + 2x^2 + 28x + 4x^2 + 18x - 8.5x + 59.5 - 34 = 0
Combining like terms:
22.5x + 25.5 = 0
-22.5x = 25.5
Dividing by -22.5:
x = -25.5 / 22.5
Simplifying:
x ≈ -1.133
Now, we discard this negative value for x since it doesn't make sense in our context.
So, the value of x is approximately 1.133.
Now, we can substitute this value of x back into the volume equation to find the dimensions of the box:
Length of the base = 8.5 - 2x
Width of the base = 14 - 2x
Height of the box = x
Substituting the value of x:
Length of the base ≈ 8.5 - 2(1.133)
Width of the base ≈ 14 - 2(1.133)
Height of the box ≈ 1.133
Calculating the approximate values:
Length of the base ≈ 6.234
Width of the base ≈ 11.734
Height of the box ≈ 1.133
Therefore, the dimensions of the box with maximum volume, rounded to three decimal places, are approximately 6.234, 11.734, and 1.133.