A Candy box is made from a piece of cardboard that meaasures 11 by 7 inches. Squares of equal size will be cut out of each corner. The sides will then be folded up to form a rectangular box. What size square should be cut from each corner to obtain maximum volume?
Let squares of size x" be cut from the corners.
Volume of (open) box
=V(x)
=height*length*width
=x(11-2x)(7-2x)
=4x^3-36x^2+77x
For maximum (or minimum), equate derivative to zero:
dV(x)/dx = 12x²-72x+77=0
Solve for x to get
x=3±(√93)/6
=1.39 or 4.61 (approximately)
4.61 is clearly not a feasible solution (because 2*4.61 > 7") and will be rejected.
So the cut-outs will be squares of 1.39" (approximately).
Now verify that the solution so obtained is a maximum by ensuring that d²V(x)/dx² < 0:
d²V(x)/dx² = 24x-72 = -38.6 <0 OK.
To determine the size of the square that should be cut from each corner to obtain the maximum volume of the candy box, follow these steps:
Step 1: Visualize the scenario
Draw a diagram or visualize a rectangular piece of cardboard measuring 11 inches by 7 inches.
Step 2: Identify the side lengths of the resulting box
After cutting squares of equal size from each corner, the length of the resulting box will be 11 - 2x (as 2 squares will be cut from each side) and the width will be 7 - 2x.
Step 3: Calculate the volume formula
The volume of a rectangular box is determined by multiplying its length, width, and height. In this scenario, the height will be equal to the side length of the square cut from each corner.
Volume = length * width * height
Step 4: Write the volume equation
Using the information from Step 2 and Step 3, the volume equation becomes:
V(x) = (11 - 2x) * (7 - 2x) * x
= x * (11 - 2x) * (7 - 2x)
Step 5: Simplify the volume equation
Expand the expression:
V(x) = (77 - 22x - 14x + 4x^2) * x
= (4x^2 - 36x + 77) * x
= 4x^3 - 36x^2 + 77x
Step 6: Find the derivative of the volume equation
Differentiate the volume equation to find where the maximum or minimum value occurs:
V'(x) = 12x^2 - 72x + 77
Step 7: Set the derivative equal to zero and solve for x
To find the critical points where the maximum or minimum occurs:
12x^2 - 72x + 77 = 0
Unfortunately, the solutions to this quadratic equation are not nice whole numbers. Using the quadratic formula, x is approximately 1.5028 and 3.1645.
Step 8: Determine the valid solution
Since x represents the side length of a square, it cannot be negative or larger than the side length of the cardboard. Thus, the valid solution is x ≈ 1.5028.
Therefore, to obtain the maximum volume, squares of approximately 1.5028 inches should be cut from each corner of the cardboard.
To find the size of the square that should be cut from each corner to obtain the maximum volume of the candy box, we need to follow these steps:
1. Visualize the problem: Imagine a piece of cardboard measuring 11 by 7 inches.
2. Determine the variables: Let's define the size of the squares that will be cut from each corner as "x". This means that the length and width of the box will be reduced by 2x (since there are two corners per side).
3. Calculate the dimensions: After cutting squares of size "x" from each corner, the dimensions of the rectangular box will be (11 - 2x) by (7 - 2x). The height of the box will be "x".
4. Calculate the volume: The volume of the box can be determined by multiplying its length, width, and height. In this case, the volume can be represented as V = (11 - 2x)(7 - 2x)x.
5. Simplify the volume equation: Expanding the equation, we have V = 4x^3 - 36x^2 + 77x.
6. Find the derivative: To find the maximum volume, we need to find the critical points of the volume equation. Taking the derivative of the volume equation with respect to x, we get dV/dx = 12x^2 - 72x + 77.
7. Solve for critical point: Setting the derivative equal to zero, we solve the equation 12x^2 - 72x + 77 = 0 to find the critical points (values of x).
8. Find the maximum value: Use the second derivative test or graphing to determine if the critical point found in the previous step is a maximum or minimum. In this case, the second derivative is positive, indicating it is a local minimum.
9. Evaluate the solution: Since we are trying to maximize the volume, we are looking for a maximum value. Therefore, the local minimum found in step 8 will be our solution.
Finally, it is important to note that the units in this problem are inches, so the size of the square that should be cut from each corner to obtain the maximum volume is the value of "x" obtained in step 7.