Pick any exercise in Chapter 12 of the calculus textbook. Post the solution or your attempt. Discuss any issues you might have had. What were the most valuable elements you took away from this course?
\[f(x,y) = x^{2} + 4x + y^{2} + -9y + 3xy\]
We compute the first and second partial derivatives of \(f\): \[ \begin{array} ff_{x}(x,y) = 2x + 1 + 3y & f_{y}(x,y) = 2y - 9 + 3x \\ \end{array} \]
Each derivative is always defined. To find the critical points - at which both \(f_{x}\) and \(f_{y}\) are \(0\) or undefined - we find the first derivative and solve the system of linear equations.
\[ \begin{align} f_{x} &= 2x + 1 + 3y = 0 \\ & 2x = -3y - 1 \\ & x = \frac{-3y - 1}{2} \\ \\ f_{y} &= 2y - 9 + 3x = 0 \\ & 2y = -3(x) + 9 \\ & 2y = -3(\frac{-3y - 1}{2}) + 9 \\ & 4y = 9y + 3 + 18 \\ \\ & y = -\frac{21}{5} \\ \\ &x = -3(-\frac{21}{5}) - 1 = \frac{58}{5} \end{align} \] The solution to the system of equations is \(x = 11\frac{3}{5}\) and \(y = -4\frac{1}{5}\).
We then apply the second derivative test:
\[ \begin{array} ff_{xx} = 2 & f_{yy} = 2 & f_{xy} = 3\\ \\ \end{array} \\ \begin{align} D(x,y) &= ff_{xx}(x_{0}, y_{0}) f_{yy}(x_{0}, y_{0}) - f_{xy}^2(x_{0}, y_{0})\\ &= (2)(2) - (3)^{2} = -5 \\ \end{align} \]
AS \(D = -5 < 0\), \((\frac{58}{5}, -\frac{21}{5})\) is a saddle point.
\[f(x,y) = x^{2} + 2x + y^{2} + 2y, \text{constrained to region bounded by } y = x^{2} + y^{2} = 4\]
We are looking for the intersection of f with a circle centered at the origin with radius of 2, which we’ll call \(g(x)\).
\[ \begin{align} g(x) &= x^{2} + y^{2} - 4 \\ & y^{2} = -x^{2} + 4 \\ & y = \sqrt{-x^{2} + 4} \\ \\ f(x,y) &= x^{2} + 2x + y^{2} + 2y \\ &= x^{2} + 2x + ({\sqrt{-x^{2} + 4}})^{2} + 2(\sqrt{-x^{2} + 4}) \\ &= x^{2} + 2x + (-x^{2} + 4) + 2(\sqrt{-x^{2} + 4}) \\ &= 2x + 4 + 2(\sqrt{-x^{2} + 4}) \\ \\ f'(x,y) &= 2 + 2(\frac{1}{2})(\frac{1}{\sqrt{-x^{2} + 4}}) = 2 + \frac{1}{\sqrt{-x^{2} + 4}} \\ & 2 + \frac{1}{\sqrt{-x^{2} + 4}} = 0 \\ & \sqrt{-x^{2} + 4} = -\frac{1}{2} \\ & -x^{2} + 4 = \frac{1}{4} \\ & x = \sqrt \frac{15}{4} \\ & y^{2} = -x^{2} + 4 \\ & y^{2} = - \sqrt \frac{15}{4}^{2} + 4 \\ & y = \frac{15}{4} + 4 = \frac{31}{4} \\ \end{align} \]
There’s a single critical point at \(g(\sqrt \frac{15}{4}, \frac{31}{4})\) on the intersection of with the circle. To determine whether this is a minimum or a maximum, we perform the second derivative test.
\[ g'(x,y) = 2 + \frac{1}{\sqrt{-x^{2} + 4}} \\ \begin{array} fg_{xx} = \frac{1}{(\sqrt{-x^{2} + 4)^{3}}} & g_{yy} = 0 & f_{xy} = 0\\ \\ \end{array} \\ \begin{align} D(x,y) &= ff_{xx}(x_{0}, y_{0}) f_{yy}(x_{0}, y_{0}) - f_{xy}^2(x_{0}, y_{0})\\ &= (\frac{1}{(\sqrt{-x^{2} + 4)^{3}}})(0) - (0)^{2} = 0 \\ \end{align} \]
As \(D = 0\), the test is inconclusive. It’s possible derivatives or system of equations algebra introduced an error.