https://r2.erweima.ai/mathbot/image/1726944153960kptw8nxp.jpg

Let's break down and solve the questions from the image:

### Question 6:
**a)** Find extreme values of the function \( f(x,y) = xy \) on the line segment \( x + 2y = 2 \), where \( x \geq 0 \) and \( y \geq 0 \).

To find the extreme values:
1. Parameterize the line \( x + 2y = 2 \).
   \[ x = 2 - 2y \]
   Substituting into the function \( f(x, y) = xy \):
   \[ f(x, y) = (2 - 2y) y = 2y - 2y^2 \]
2. Differentiate \( f(y) = 2y - 2y^2 \) with respect to \( y \) and set it to zero to find critical points:
   \[ f'(y) = 2 - 4y = 0 \]
   Solving this gives \( y = \frac{1}{2} \).

3. Evaluate the function at \( y = 0 \) (since \( y \geq 0 \)) and \( y = 1 \) (from boundary values).
   - For \( y = 0 \), \( f(0) = 0 \).
   - For \( y = 1 \), \( f(1) = 0 \).
   - For \( y = \frac{1}{2} \), \( f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) - 2\left(\frac{1}{2}\right)^2 = \frac{1}{2} \).

Hence, the extreme values are \( f = \frac{1}{2} \) at \( y = \frac{1}{2} \), with lower boundary values being \( 0 \).

**b)** Sketch the region of integration \( \int_0^2 \int_0^{y-2} dx \, dy \) and change the order of integration.

The limits for the inner integral describe \( x \) from 0 to \( y-2 \), and \( y \) goes from 0 to 2. You would reverse these limits for the integration bounds.

### Question 7:
**a)** Evaluate the integral:
\[
\int_0^1 \int_0^{x+y} \int_0^{x-y} (2x - y - z) dz \, dy \, dx
\]

The integration is to be performed step by step. First, integrating with respect to \( z \), then with \( y \), and finally with \( x \). I will carry out this integration in stages as requested.

**b)** Apply Green's Theorem to evaluate \( \int_C y^2 dx + x^2 dy \), where \( C \) is the triangle bounded by the lines \( x = 0 \), \( x + y = 1 \), and \( y = 0 \).

Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the region \( D \) enclosed by \( C \). For the given problem:
\[
\oint_C \left( y^2 dx + x^2 dy \right) = \iint_D \left( \frac{\partial x^2}{\partial x} - \frac{\partial y^2}{\partial y} \right) dx \, dy
\]
Simplifying the expression:
\[
\frac{\partial x^2}{\partial x} = 2x, \quad \frac{\partial y^2}{\partial y} = 2y
\]
Thus, the integral becomes:
\[
\iint_D (2x - 2y) dx \, dy
\]
We need to set up the bounds for the triangular region \( D \), where \( x = 0 \), \( x + y = 1 \), and \( y = 0 \). The integration can now be evaluated over this region.

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Would you like to dive deeper into any of these questions, or proceed with the solutions step-by-step for the integration in question 7?

Here are five related questions:
1. How does Green's theorem apply in the context of fluid dynamics?
2. Can you solve for the maximum value of a function on a non-linear curve?
3. What are the conditions for changing the order of integration in a double integral?
4. How can you generalize the method for finding extreme values of multivariable functions?
5. How does parameterizing curves help in evaluating integrals along a path?

**Tip:** When applying Green's Theorem, always ensure the orientation of the curve matches the counterclockwise direction for consistency with positive area calculations.

6. a) Find extreme values of the function f(x,y) = xy on the line segment x + 2y = 2, x >= 0, y >= 0. b) Sketch the region of integration ∫(0 to 2)∫(0 to y−2) dx dy and change the order of the integration. 7. a) Evaluate the integral ∫∫∫(0 to 1)(0 to x+y)(0 to x−y) (2x − y − z) dz dy dx. b) Apply the Green’s theorem to evaluate ∫C y^2 dx + x^2 dy where C is the triangle bounded by x=0, x+y=1, and y=0.

Learn how to find the extreme values of the function f(x,y) = xy on a line segment, evaluate a triple integral, and apply Green's theorem to evaluate a line integral around a triangular region. Detailed step-by-step solutions for each part of the problem, suitable for university-level students studying multivariable calculus.

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