Determine the maximum value
of y=3x2−x
3
A. 0
B. 2
C. 4
D. 6
To find the maximum value of y, we can take the derivative of y with respect to x and find the critical points.
dy/dx = 6x - 3x^2
Setting the derivative equal to zero, we have:
6x - 3x^2 = 0
Factoring out x, we get:
x(6 - 3x) = 0
Setting each factor equal to zero, we have two critical points:
x = 0 and x = 2
Now, we can find the value of y at these critical points.
For x = 0:
y = 3(0)^2 - 0^3 = 0
For x = 2:
y = 3(2)^2 - 2^3 = 12 - 8 = 4
Comparing the y values at these critical points, we see that the maximum value of y is 4.
Therefore, the answer is C. 4.
To determine the maximum value of the equation y = 3x^2 - x^3, we need to find the point at which the derivative equals zero.
First, we find the derivative of y with respect to x:
dy/dx = 6x - 3x^2
Next, we set the derivative equal to zero and solve for x:
6x - 3x^2 = 0
Factoring out x, we get:
x(6 - 3x) = 0
Setting each factor equal to zero, we find two possible solutions:
x = 0 or 6 - 3x = 0
Simplifying the second equation, we have:
6 - 3x = 0
3x = 6
x = 2
Now that we have the critical points x = 0 and x = 2, we need to evaluate the value of y at each point to determine which one yields the maximum.
For x = 0:
y = 3(0)^2 - (0)^3
y = 0 - 0
y = 0
For x = 2:
y = 3(2)^2 - (2)^3
y = 3(4) - 8
y = 12 - 8
y = 4
Therefore, the maximum value of y is 4.
The correct answer is C. 4