a rectangular sheet of cardboard size 5' by 8' is to be used to make an open box by cutting out the four corners. What is the largest volume and be sure to explain how you found the MAXIMUM volume please and Thank You!
Are you allowed to use differential calculus?
Write an expression for box volume V in terms of the length of the corner cutouts, x.
V = (5-2x)(8-2x)*x
= 4x^3 -26x^2 +40x
Solve for x where V'(x) = dV/dx = 0
dV/dx = 12 x^2 -52x + 40 = 0
3x^2 -13x +10 = 0
x = (1/6)[13 +/- 7] = 1 and 3.333
The second root is not possible, because the removed material would exceed the length of the 5' side.
The largest volume is what you get with x = 1 foot.
Vmax = 18 ft^3
To find the maximum volume of the open box, we need to determine the dimensions that will result in the largest possible volume.
Let's start by visualizing the process. We have a rectangular sheet with dimensions 5' by 8'. To create an open box, we need to cut out squares from each corner. Let's assume we cut x feet from each corner. This will leave us with a box with dimensions (5 - 2x) by (8 - 2x) by x.
The volume of the box can be calculated by multiplying its length, width, and height. Thus, the volume V can be expressed as V = (5 - 2x) * (8 - 2x) * x.
To find the maximum volume, we use optimization techniques by differentiating the volume equation and setting the derivative equal to zero.
We have V = (5 - 2x) * (8 - 2x) * x.
Expanding this expression, we get V = 4x^3 - 26x^2 + 40x.
Now, let's differentiate the volume equation with respect to x: dV/dx = 12x^2 - 52x + 40.
Setting dV/dx equal to zero and solving for x, we have:
12x^2 - 52x + 40 = 0.
Now we can solve this quadratic equation. Factoring out a common factor of 4, we get:
4(3x^2 - 13x + 10) = 0.
Simplifying further, we have:
3x^2 - 13x + 10 = 0.
Factoring the quadratic equation, we have:
(3x - 2)(x - 5) = 0.
Setting each factor equal to zero, we find two possible solutions:
3x - 2 = 0, which gives x = 2/3
x - 5 = 0, which gives x = 5.
Since we want to maximize the volume, we discard the solution x = 2/3 (as it would result in a negative dimension) and consider x = 5 as the value we need.
Therefore, the largest volume of the open box is obtained when we cut squares of dimension 5 feet from each corner.
To find the maximum volume, substitute x = 5 into the volume equation:
V = (5 - 2x) * (8 - 2x) * x.
V = (5 - 2(5)) * (8 - 2(5)) * 5.
V = (5 - 10) * (8 - 10) * 5.
V = (-5) * (-2) * 5.
V = 50 cubic feet.
Therefore, the largest volume of the open box is 50 cubic feet, achieved by cutting squares with dimensions of 5 feet from each corner.