V(x)=x(10−2x)(12−2x)
. Find the maximum volume of a box that can be created from the piece of paper.
To find the maximum volume of the box, we need to maximize the function V(x) by finding its critical points.
First, we find the derivative of V(x) with respect to x:
V'(x) = (10 - 2x)(12 - 2x) + x(-2)(12 - 2x) + x(10 - 2x)(-2)
= (10 - 2x)(12 - 2x) - 2x(12 - 2x) - 2x(10 - 2x)
= (10 - 2x)(12 - 2x) - 24x + 4x^2 - 20x + 4x^2
= (10 - 2x)(12 - 2x) + 8x^2 - 44x
To find the critical points, we set V'(x) = 0 and solve for x:
(10 - 2x)(12 - 2x) + 8x^2 - 44x = 0
(10 - 2x)(12 - 2x) = 44x - 8x^2
120 - 24x - 20x + 4x^2 = 44x - 8x^2
4x^2 + 44x - 24x - 20x - 120 = 0
4x^2 + 0x - 120 = 0
x^2 - 30 = 0
(x - √30)(x + √30) = 0
Setting each factor equal to zero:
x - √30 = 0 or x + √30 = 0
x = √30 or x = -√30 (rejected since x should be positive in this context)
Therefore, the critical point is x = √30.
To determine if this critical point is a maximum or minimum, we evaluate the second derivative:
V''(x) = 8 - 44 = -36
Since V''(√30) = -36, which is negative, this means that x = √30 is a maximum point.
We can plug x = √30 back into the original equation for V(x) to find the maximum volume:
V(√30) = √30( 10 - 2√30)(12 - 2√30).
Approximating to three decimal places, the maximum volume of the box is V(√30)= 51.962 cubic units.
To find the maximum volume of a box that can be created from the given piece of paper, we need to find the value of x that maximizes the function V(x) = x(10 - 2x)(12 - 2x).
Step 1: Expand the equation
V(x) = x(10 - 2x)(12 - 2x)
= x(120 - 20x - 24x + 4x^2)
= x(4x^2 - 44x + 120)
Step 2: Simplify the equation
V(x) = 4x^3 - 44x^2 + 120x
Step 3: Take the derivative of V(x) with respect to x
V'(x) = 12x^2 - 88x + 120
Step 4: Set V'(x) equal to 0 to find critical points
12x^2 - 88x + 120 = 0
Step 5: Solve the quadratic equation
Using factoring or the quadratic formula, we find that x = 1 or x = 10/3.
Step 6: Determine the second derivative of V(x)
V''(x) = 24x - 88
Step 7: Plug in the values of x from Step 5 into V''(x)
V''(1) = -64
V''(10/3) = 8/3
Since V''(1) is negative and V''(10/3) is positive, the critical point x = 10/3 corresponds to a local maximum.
Thus, the maximum volume of the box that can be created from the piece of paper is V(10/3).
You can calculate the value of V(10/3) to find the specific volume of the box.
To find the maximum volume of the box that can be created, we need to maximize the volume function V(x) = x(10 - 2x)(12 - 2x), where x represents the length of the side of the square that is cut out from each corner of the paper.
First, let's expand the function:
V(x) = x(10 - 2x)(12 - 2x)
= x(120 - 24x - 20x + 4x^2)
= x(4x^2 - 44x + 120)
= 4x^3 - 44x^2 + 120x
To find the maximum volume, we need to take the derivative of the volume function with respect to x and set it equal to zero:
V'(x) = 12x^2 - 88x + 120 = 0
Now, we need to solve this quadratic equation for x. We can either factor it or use the quadratic formula. Let's use the quadratic formula to find the values of x:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 12, b = -88, and c = 120. Applying the quadratic formula, we have:
x = (-(-88) ± √((-88)^2 - 4(12)(120)))/(2(12))
x = (88 ± √(7744 - 5760))/(24)
x = (88 ± √1984)/(24)
Now we can simplify further:
x = (88 ± 44√2)/(24)
The two solutions are:
x₁ = (88 + 44√2)/(24)
x₂ = (88 - 44√2)/(24)
To determine which value of x gives the maximum volume, we need to test these values. Plug each value of x back into the volume function V(x) and compare the results. The value of x that gives the greatest volume will correspond to the maximum volume of the box.
V(x₁) = 4(x₁)^3 - 44(x₁)^2 + 120(x₁)
V(x₂) = 4(x₂)^3 - 44(x₂)^2 + 120(x₂)
Calculate these values and compare them to find the maximum volume.
Note: Depending on the range of x-values that are realistic for this problem, you may need to check the endpoints of that range as well to ensure you find the absolute maximum volume.