1. A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume?
so I did V=lhw
and found l=4-2h and w=3-2h
I also distributed them in the equation
I found V= and found the derivative
v'=12-28h+12h^2 and I set it equal to zero
from there...I don't know whay to do.
v= h (4-2h)(3-2h) = h (12 - 14 h + 4 h^2)
v = 12 h - 14 h^2 + 4 h^3
0 = 12 - 28 h + 12 h^2 agreed now factor 4 out
3 h^2 -7 h + 3 = 0
h = [ 7 +/- sqrt(49 - 36) ]/6
= [ 7 +/- sqrt(13)]/6
= 1.77 or .566
1.77 is more than half the width of 3 so is no good
use h = 0.566
To find the dimensions of the box with the largest volume, you have correctly set up the volume equation and found the derivative of volume with respect to one of the dimensions. Let's continue from there.
So, you have found the derivative of the volume equation, which is v' = 12 - 28h + 12h^2. To find the value of h at which the derivative is zero, you need to set v' equal to zero and solve for h.
12 - 28h + 12h^2 = 0
Now, we need to solve this equation for its roots. One way to solve quadratic equations like this is by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 12, b = -28, and c = 12. Substituting these values into the quadratic formula, we get:
h = (-(-28) ± √((-28)^2 - 4*12*12)) / (2*12)
h = (28 ± √(784 - 576)) / 24
h = (28 ± √208) / 24
Now, let's simplify the expression inside the square root:
√208 = √(16 * 13) = 4√13
Therefore, the equation for h becomes:
h = (28 ± 4√13) / 24
Now, let's simplify further by dividing both the numerator and denominator by 4:
h = (7 ± √13) / 6
Now, we have two possible values for h: (7 + √13) / 6 and (7 - √13) / 6.
Since we are looking for the dimensions of the box with the largest volume, we need to choose the value of h that maximizes the volume. To do this, we need to consider the range of possible values for h.
The dimensions of the original sheet are 3ft by 4ft, so the maximum possible value for h is the smaller of the two sides, which is 3ft.
By evaluating the two possible values for h, we find:
(7 + √13) / 6 ≈ 2.23
(7 - √13) / 6 ≈ 0.47
Since 2.23 is greater than 0.47, we can discard the smaller value as it exceeds the maximum possible value for h.
Therefore, the value of h that maximizes the volume is approximately 2.23.
To find the corresponding values for the other dimensions, we can substitute this value of h back into the expressions for l and w:
l = 4 - 2h
l = 4 - 2(2.23)
l ≈ 4 - 4.46
l ≈ -0.46
Since the length of a box cannot be negative, we can discard this result.
w = 3 - 2h
w = 3 - 2(2.23)
w ≈ 3 - 4.46
w ≈ -1.46
Again, since the width of a box cannot be negative, we discard this result as well.
In conclusion, based on the calculations, we have found that there is no box with integer dimensions that can be formed from a sheet of cardboard 3 ft. by 4 ft. that has the largest volume.