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Answer :
Final answer:
To maximize the volume of the resulting box, we need to set up an equation using the dimensions of the original cardboard and the size of the squares cut out. We can then take the derivative of the volume equation, set it equal to zero, and solve for the size of the squares cut out. Finally, we can substitute the value of x back into the dimensions equations to find the dimensions of the resulting box.
Explanation:
To find the dimensions that maximize the volume of the resulting box, we need to set up an equation in terms of the length L, width W, and height H. From the given information, we know that after cutting squares out of each corner and folding up the sides, the length L of the resulting box will be the length of the original cardboard minus twice the size of the squares cut out. Similarly, the width W of the resulting box will be the width of the original cardboard minus twice the size of the squares cut out.
Let's denote the size of the squares cut out as x. So, L = 20 - 2x and W = 12 - 2x.
Now, we can express the volume of the resulting box in terms of x. The height H of the box will be equal to x. Therefore, the volume V of the box is V = LWH = (20 - 2x)(12 - 2x)x. To find the maximum volume, we can take the derivative of V with respect to x, set it equal to zero, and solve for x.
After finding the value of x, we can substitute it back into the expressions for L, W, and H to obtain the dimensions of the resulting box that maximizes the volume.
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Answer:
Box Dimensions:
L = 15.15 ul
W = 7.15 ul
h = x = 2.43 ul
V(max) = 263.22 cu
Step-by-step explanation:
We call x the length of the square to be cut in the corners then:
Are of the base of the box is:
(20 - 2*x) is the future length of the box and
(12 - 2*x) will be the width
The heigh is x then the volume of the box is:
V = ( 20 - 2*x )* ( 12 - 2*x ) * h
And the volume as a function of x is:
V(x) = ( 20 - 2*x) * ( 12 - 2*x ) * x or V(x) = (240 -40*x -24*x + 4*x²) * x
V(x) = 240*x - 64*x² + 4*x³
Taking derivatives on both sides of the equation we get:
V´(x) = 240 - 128*x + 12*x²
V´(x) = 0 240 - 128*x + 12*x² = 0 or 60 - 32*x + 3*x²
3*x² - 32*x + 60 = 0
Solving:
x₁,₂ = 32 ± √ (32)² - 4*3*60 ]/ 2*3
x₁,₂ = 32 ± √ 1024 - 720 )/6
x₁,₂ = ( 32 ± √ 304 )/6
x₁,₂ = ( 32 ± 17.44 )/6
x₁ = 8.23 ( we dismiss this solution because is not feasible 2*x > 12
x₂ = 2.43 u.l ( units of length)
Then
L = 20 - 2*x L = 20 - 4.85 L = 15.15 ul
W = 12 - 2*x W = 12 - 4.85 W = 7.15 ul
h = 2.43 ul
V = 2.43*7.15*15.15 cubic units
V = 263.22 cu
To see if when x = 2.43 function V has a maximum we go to the second derivative
V´´(x) = - 128 + (24)*2.43
V´´(x) = - 69.68 as V´´(x) < 0 then we have a maximum for V(x) in the point x = 2.43
