Math Problem Statement

Constraint: x>30 and x<6666

Solution

Given the constraint x>30x > 30 and x<6666x < 6666, we need to find the maximum value of the function f(x)=(x30)(12000021x+x2)f(x) = (x-30)(120000-21x+x^2) within this interval. We will use calculus and the derivative test to find the critical points within this range.

First, we previously found the first derivative:

f(x)=3x2102x+120630f'(x) = 3x^2 - 102x + 120630

We set the derivative equal to zero to find the critical points:

3x2102x+120630=03x^2 - 102x + 120630 = 0

Now let's solve this quadratic equation:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a=3a = 3, b=102b = -102, and c=120630c = 120630.

x=102±1040414475606x = \frac{102 \pm \sqrt{10404 - 1447560}}{6}

Since the discriminant is negative (Δ=b24ac=1437156\Delta = b^2 - 4ac = -1437156), there are no real solutions to the quadratic equation.

In this case, we need to evaluate the function at the boundaries of the interval x>30x > 30 and x<6666x < 6666:

  1. Evaluate f(x)f(x) at x=30x = 30:

f(30)=(3030)(12000021(30)+302)f(30) = (30-30)(120000 - 21(30) + 30^2) f(30)=0f(30) = 0

  1. Evaluate f(x)f(x) at x=6666x = 6666:

f(6666)=(666630)(12000021(6666)+66662)f(6666) = (6666-30)(120000 - 21(6666) + 6666^2) f(6666)=6636(12000021(6666)+66662)f(6666) = 6636(120000 - 21(6666) + 6666^2) f(6666)=6636(120000139986+44435556)f(6666) = 6636(120000 - 139986 + 44435556) f(6666)=6636(44425570)f(6666) = 6636(44425570)

Since evaluating this directly gives a very large number, let's consider the nature of the function:

Given the cubic polynomial f(x)=x351x2+120630x3600000f(x) = x^3 - 51x^2 + 120630x - 3600000, we can approximate where the maximum might lie by checking intermediate values within the interval:

  1. Evaluate f(x)f(x) at a few intermediate points:

Let's choose x=3000x = 3000 and x=5000x = 5000:

f(3000)=(300030)(12000021(3000)+30002)f(3000) = (3000-30)(120000 - 21(3000) + 3000^2) f(3000)=2970(12000063000+9000000)f(3000) = 2970(120000 - 63000 + 9000000) f(3000)=2970(8883000)f(3000) = 2970(8883000)

f(5000)=(500030)(12000021(5000)+50002)f(5000) = (5000-30)(120000 - 21(5000) + 5000^2) f(5000)=4970(120000105000+25000000)f(5000) = 4970(120000 - 105000 + 25000000) f(5000)=4970(25025000)f(5000) = 4970(25025000)

  1. Compare these values:

The maximum of the function seems to be growing with xx within the interval. Given this behavior, we hypothesize the maximum might be at the upper boundary x=6666x = 6666.

Thus, we find that x=6666x = 6666 maximizes the function f(x)=(x30)(12000021x+x2)f(x) = (x-30)(120000-21x+x^2) within the given constraint 30<x<666630 < x < 6666.

Would you like more details on any part of this solution or have any specific questions?

Here are some relative questions for further exploration:

  1. What practical application could this function represent?
  2. How can we interpret the maximum value in a real-world context?
  3. Can we further refine the interval to get a more precise maximum value?
  4. How would the result change if the interval was different?
  5. What numerical methods can we use to verify the maximum value?
  6. How does the behavior of the polynomial influence our solution?
  7. What would happen if the constraints were extended or reduced?
  8. Can this approach be generalized to other similar problems?

Tip: When dealing with constrained optimization problems, always check the function's values at the boundaries of the interval as well as any critical points within the interval.

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Math Problem Analysis

Mathematical Concepts

Calculus
Quadratic Equations
Critical Points

Formulas

Quadratic formula
Derivative formulas

Theorems

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Suitable Grade Level

Advanced Mathematics