Chocolate Box Company is going to make open-topped boxes out of 3 × 11-inch rectangles of cardboard by cutting squares out of the corners and folding up the sides. What is the largest volume box it can make this way? (Round your answer to one decimal place.)
My work
(11-2x)(3-2x)
(4x^2-28x+33)x
4x^3-28x^2+33x
12x^2 -56x+33 <---- where i am stuck
maximum volume is where the derivative is zero.
12x^2-56x+33 = 0
so, just use the quadratic formula to find where that is.
To find the largest volume box, we need to maximize the volume function. The volume of the box is given by the equation V = length × width × height.
In this case, the length and width of the box are determined by the dimensions of the cardboard rectangle after cutting squares out of the corners. Let's assume the length of the box is (11 - 2x) inches and the width is (3 - 2x) inches, where x is the length of each side of the square cut out of the corners.
Since the box is open-topped, the height of the box will be the length of each side of the square cut out of the corners, which is also x inches.
So, V = (11 - 2x)(3 - 2x)(x).
To find the largest volume, we need to find the value of x that maximizes this function.
Next, let's simplify the volume function by multiplying:
V = (11 - 2x)(3 - 2x)(x)
= (33 - 22x + 4x^2 - 6x + 4x^2 - 2x^3)(x)
= (33 - 28x + 8x^2 - 2x^3)(x)
= -2x^4 + 8x^3 + 8x^2 - 28x^2 + 33x
= -2x^4 + 8x^3 - 20x^2 + 33x
Now, we have the volume function as a polynomial equation: V = -2x^4 + 8x^3 - 20x^2 + 33x.
To find the maximum volume, we can differentiate this function with respect to x and set the derivative equal to zero:
V' = -8x^3 + 24x^2 - 40x + 33 = 0.
Now, we need to solve this equation to find the critical points where the derivative is equal to zero.
At this point, the equation becomes more complicated to solve algebraically. You can use numerical methods or graphing calculators to find the values of x that satisfy the equation V' = 0. One common numerical method is called Newton's method.
Alternatively, you can use various software programs or online calculators that can solve polynomial equations to find the values of x where the derivative is zero.
Once you find those critical points, substitute them back into the volume function, V = -2x^4 + 8x^3 - 20x^2 + 33x, to determine the largest volume.
Note: Without the specific values of x resulting from solving the equation V' = 0, it's not possible to determine the exact largest volume in this particular problem.