Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Th...
F(x) = min (2x
2, 52 – 5x)
The maximum possible value of this function will be attained at the point in which 2x2 is equal to 52 – 5x.
2x2 = 52 − 5x
2x2 + 5x − 52 = 0
(2x + 13) (x-4) = 0
=>x = -13 / 2 or x = 4 It has been given that is a positive real number. Therefore, we can eliminate the case x = -13 / 2 . x = 4 is the point at which the function attains the maximum value. 4 is not the maximum value of the function. Substituting x=4 in the original function, we get, 2x2 = 2* 42 = 32.
f(x) = 32.
Therefore, 32 is the right answer.
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Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Th...
Explanation:
To find the maximum possible value of f(x), we need to determine the value of x that results in the maximum value of f(x).
Step 1: Finding the critical points
To find the critical points of f(x), we need to determine the values of x where the derivative of f(x) is equal to zero.
Let's find the derivative of f(x) with respect to x:
f'(x) = d/dx(min(2x^2, 52 - 5x))
To find the derivative of min(2x^2, 52 - 5x), we need to consider two cases:
Case 1: When 2x^2 is smaller than 52 - 5x
In this case, the derivative of f(x) with respect to x is equal to the derivative of 2x^2:
f'(x) = d/dx(2x^2) = 4x
Case 2: When 2x^2 is greater than or equal to 52 - 5x
In this case, the derivative of f(x) with respect to x is equal to the derivative of (52 - 5x):
f'(x) = d/dx(52 - 5x) = -5
Step 2: Analyzing the critical points
To find the critical points, we set the derivatives equal to zero and solve for x:
Case 1: When 2x^2 is smaller than 52 - 5x:
4x = 0
x = 0
Case 2: When 2x^2 is greater than or equal to 52 - 5x:
-5 = 0
No solutions exist for this case.
Step 3: Determining the maximum value of f(x)
To determine the maximum value of f(x), we need to evaluate f(x) at the critical points and the endpoints of the domain.
Case 1: Evaluate f(x) at x = 0:
f(0) = min(2(0)^2, 52 - 5(0)) = min(0, 52) = 0
Case 2: Evaluate f(x) at the endpoint of the domain:
Since x is a positive real number, there is no endpoint of the domain.
Conclusion:
Therefore, the maximum possible value of f(x) is 0, which contradicts the given answer of 32. It is possible that there is an error in the given information or question.
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