Math Problem Statement

Find the absolute maximum and minimum values of f(x,y)=x^3+3y^2+4 on the set D where D is the closed region bounded by y=0 and y=5−5x^2. Along y=5−5x2 , f can be expressed as a function of one variable List all the points on this side of the boundary which could potentially be the absolute minimum or maximum on D .

Solution

We are given the function f(x,y)=x3+3y2+4f(x, y) = x^3 + 3y^2 + 4, and we are asked to find the absolute maximum and minimum values of this function on the closed region DD, which is bounded by the curves y=0y = 0 and y=55x2y = 5 - 5x^2.

Steps to find the absolute maximum and minimum:

  1. Analyze the boundary of the region DD:

    • DD is bounded by y=0y = 0 (the x-axis) and y=55x2y = 5 - 5x^2.
    • Along y=55x2y = 5 - 5x^2, we can express f(x,y)f(x, y) as a function of xx, which allows us to reduce the problem to a single-variable function.
  2. Express f(x,y)f(x, y) on the curve y=55x2y = 5 - 5x^2: Substitute y=55x2y = 5 - 5x^2 into the function f(x,y)f(x, y):

    f(x,55x2)=x3+3(55x2)2+4f(x, 5 - 5x^2) = x^3 + 3(5 - 5x^2)^2 + 4

    Simplifying this expression: f(x,55x2)=x3+3(2550x2+25x4)+4f(x, 5 - 5x^2) = x^3 + 3(25 - 50x^2 + 25x^4) + 4 f(x,55x2)=x3+75150x2+75x4+4f(x, 5 - 5x^2) = x^3 + 75 - 150x^2 + 75x^4 + 4 f(x,55x2)=75x4150x2+x3+79f(x, 5 - 5x^2) = 75x^4 - 150x^2 + x^3 + 79

    Now we need to find the critical points of this function within the range x[1,1]x \in [-1, 1], because y=55x2y = 5 - 5x^2 requires that x21x^2 \leq 1, meaning x[1,1]x \in [-1, 1].

  3. Find the derivative of f(x,55x2)f(x, 5 - 5x^2) to locate the critical points: f(x)=ddx(75x4150x2+x3+79)f'(x) = \frac{d}{dx}\left( 75x^4 - 150x^2 + x^3 + 79 \right) Applying the power rule: f(x)=300x3300x+3x2f'(x) = 300x^3 - 300x + 3x^2 f(x)=3x(100x2100+x)f'(x) = 3x(100x^2 - 100 + x)

  4. Solve for critical points by setting f(x)=0f'(x) = 0: 3x(100x2100+x)=03x(100x^2 - 100 + x) = 0 This equation has two factors:

    • x=0x = 0
    • 100x2100+x=0100x^2 - 100 + x = 0

    Solve 100x2+x100=0100x^2 + x - 100 = 0 using the quadratic formula: x=1±124(100)(100)2(100)=1±1+40000200=1±40001200x = \frac{-1 \pm \sqrt{1^2 - 4(100)(-100)}}{2(100)} = \frac{-1 \pm \sqrt{1 + 40000}}{200} = \frac{-1 \pm \sqrt{40001}}{200} The roots are approximately:

    \quad \text{and} \quad x \approx \frac{-1 - 200.0025}{200} \approx -1.00001$$
  5. Check the critical points and endpoints within the domain [1,1][-1, 1]:

    • The critical points within the interval are x1x \approx 1 and x1x \approx -1, along with x=0x = 0.
    • The endpoints of the domain are x=1x = -1 and x=1x = 1.

Potential points on the boundary:

From the above analysis, the points on the boundary where the function might achieve its absolute minimum or maximum are:

  • x=1x = -1
  • x=0x = 0
  • x=1x = 1

Conclusion:

These are the points on the boundary that could potentially be the absolute minimum or maximum for the function on the region DD. Next, we would evaluate the function f(x,55x2)f(x, 5 - 5x^2) at these points to determine which gives the maximum and minimum values.

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Math Problem Analysis

Mathematical Concepts

Calculus
Multivariable Functions
Optimization
Critical Points
Boundary Analysis

Formulas

f(x, y) = x^3 + 3y^2 + 4
f(x, y) | y=5-5x^2 = 75x^4 - 150x^2 + x^3 + 79
f'(x) = 3x(100x^2 - 100 + x)

Theorems

Extreme Value Theorem
Critical Point Theorem

Suitable Grade Level

Grades 11-12