8. Find The Maximum Value Of F(x, Y)=x Y Subject To 3 X + y = 60. Also Find The Corresponding Point(s) (x, y)
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Introduction
In the realm of optimization problems, constrained maximization and minimization are fundamental concepts with numerous applications in economics, engineering, and mathematics. One classic problem involves finding the maximum value of a function subject to a linear constraint. In this article, we will explore the problem of maximizing the function F(x, y) = x y under the constraint 3 x + y = 60. We will determine the maximum value of the function and identify the point(s) at which this maximum occurs.
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Understanding the Problem
The Objective Function
The function we aim to maximize is:
\[ F(x, y) = x y \]
This is a bilinear function, representing the product of two variables, x and y.
The Constraint
The constraint provided is:
\[ 3 x + y = 60 \]
This is a linear equation representing a straight line in the xy-plane.
Goal
- Find the maximum value of \( F(x, y) \) subject to the constraint.
- Find the point(s) \( (x, y) \) where this maximum occurs.
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Mathematical Approach to the Problem
Step 1: Express one variable in terms of the other
Since the constraint relates x and y linearly, we can express y in terms of x:
\[ y = 60 - 3 x \]
Step 2: Rewrite the objective function in terms of a single variable
Substitute y into \( F(x, y) \):
\[ F(x) = x (60 - 3 x) \]
\[ F(x) = 60 x - 3 x^2 \]
Now, the problem reduces to maximizing \( F(x) \) over the feasible x-values.
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Finding the Maximum of the Function \( F(x) \)
Step 1: Determine the domain
Since y = 60 - 3x, for y to be real and meaningful, we consider:
\[ y \geq 0 \Rightarrow 60 - 3 x \geq 0 \Rightarrow x \leq 20 \]
Similarly, for x ≥ 0, y ≥ 0 (assuming we are in the first quadrant), we have:
\[ x \geq 0 \]
Therefore, the domain of x is:
\[ 0 \leq x \leq 20 \]
Step 2: Find critical points by differentiation
Differentiate \( F(x) \):
\[ F'(x) = 60 - 6 x \]
Set derivative to zero to find critical points:
\[ 60 - 6 x = 0 \]
\[ 6 x = 60 \]
\[ x = 10 \]
Step 3: Confirm maximum using second derivative test
Second derivative:
\[ F''(x) = -6 \]
Since \( F''(x) = -6 < 0 \), the critical point at \( x=10 \) corresponds to a local maximum.
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Calculating the Maximum Value
Step 1: Find y at \( x=10 \)
\[ y = 60 - 3 (10) = 60 - 30 = 30 \]
Step 2: Compute \( F(x, y) \) at this point
\[ F(10, 30) = 10 \times 30 = 300 \]
Thus, the maximum value of \( F(x, y) \) under the given constraint is 300.
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The Corresponding Point(s)
The point at which this maximum occurs is:
\[ (x, y) = (10, 30) \]
This point lies on the line \( 3x + y = 60 \) and satisfies the constraints.
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Additional Considerations
Boundary Points
In constrained optimization problems, it is crucial to evaluate the objective function at boundary points:
- When \( x=0 \):
\[ y = 60 - 3(0) = 60 \]
\[ F(0, 60) = 0 \times 60 = 0 \]
- When \( x=20 \):
\[ y = 60 - 3(20) = 60 - 60 = 0 \]
\[ F(20, 0) = 20 \times 0 = 0 \]
Since the maximum at the interior point \( (10, 30) \) yields \( 300 \), which is greater than the boundary values, the maximum is confirmed at \( (10, 30) \).
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Summary of Results
| Point (x, y) | Objective Function \( F(x, y) \) |
|--------------|----------------------------------|
| (0, 60) | 0 |
| (10, 30) | 300 |
| (20, 0) | 0 |
The maximum value of \( F(x, y) \) is 300, occurring at the point (10, 30).
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Practical Applications of This Optimization
This problem exemplifies the principles used in various real-life scenarios, such as:
- Profit maximization where resources are constrained.
- Production optimization in manufacturing with limited inputs.
- Economics for maximizing utility or output given resource limitations.
- Engineering design for optimal performance under constraints.
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Conclusion
In conclusion, the problem of maximizing \( F(x, y) = xy \) subject to \( 3x + y = 60 \) is effectively solved through substitution and calculus. The key steps involve expressing one variable in terms of the other, differentiating the resulting function, and analyzing critical points within the feasible domain. The maximum value obtained is 300, achieved at the point (10, 30). This example underscores the importance of mathematical techniques such as substitution, differentiation, and boundary analysis in solving constrained optimization problems.
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Additional Tips for Solving Similar Problems
- Always identify the feasible domain based on the constraints.
- Consider boundary points as potential candidates for maxima or minima.
- Use derivatives to find critical points within the domain.
- Confirm the nature of critical points using second derivative tests.
- Interpret the results within the context of the problem to ensure practical relevance.
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References
- Mathematics for Economics and Business by Ian Jacques
- Calculus: Early Transcendentals by James Stewart
- Online resources on constrained optimization and Lagrange multipliers
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By understanding and applying these methods, students and professionals can effectively tackle a wide array of optimization problems involving linear constraints and bilinear functions.
