QQuestionMathematics
QuestionMathematics
1. Optimization with Constraints (Lagrange Multipliers)
Find the maximum and minimum of
f(x,y)=x
2
+y
2
subject to the constraint
x+y= 1
about 15 hours agoReport content
Answer
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Step 1:Set up the Lagrangian
\mathcal{L}(x, y, \lambda) = x^2 + y^2 - \lambda(x + y - 1)
Construct the Lagrangian by introducing a multiplier for the constraint.
Step 2:Compute partial derivatives and set to zero
\frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda = 0 \\ \frac{\partial \mathcal{L}}{\partial y} = 2y - \lambda = 0 \\ \frac{\partial \mathcal{L}}{\partial \lambda} = -(x + y - 1) = 0
Find stationary points by setting the gradients to zero.
Step 3:Solve the system of equations
2x - \lambda = 0 \\ 2y - \lambda = 0 \\ x + y = 1
Write the equations to solve for x, y, and lambda.
Step 4:Find x and y values
2x = 2y \implies x = y \\ x + x = 1 \implies x = \frac{1}{2},\ y = \frac{1}{2}
From the equations, x equals y, and both sum to 1, so each is one-half.
Step 5:Evaluate f(x, y) at critical point
f\left(\frac{1}{2}, \frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
Plug the critical values into the original function.
Step 6:Check for maximum and minimum on the constraint
f(x, y) = x^2 + (1-x)^2 = x^2 + 1 - 2x + x^2 = 2x^2 - 2x + 1
Express f in terms of x only, using the constraint.
Step 7:Analyze the quadratic for extrema
\text{Vertex at } x = \frac{-(-2)}{2 \cdot 2} = \frac{1}{2}
The quadratic opens upwards, so the minimum is at the vertex, and maximum at the endpoints if any.
Step 8:Consider endpoints for unbounded domain
Since x and y are not bounded, as x approaches infinity or negative infinity, f(x, y) increases without bound. Thus, there is no maximum, only a minimum.
Final Answer
\text{Minimum: } \frac{1}{2} \text{ at } (\frac{1}{2}, \frac{1}{2}) \\ \text{Maximum: } \infty \text{ (unbounded)}
The minimum value is one-half at (1 / 2, 1 / 2). There is no maximum since the function increases without bound.
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