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.
V(x)=x(10−2x)(12−2x)
. Find the maximum volume of a box that can be created from the piece of paper.
1 answer