1a.) this problem was solved in lecture

1b.) You get $\nabla f = < ye^{xy},xe^{xy} >$ and $\nabla g = < 3x^2, 3y^2 >$. So your system of equations is $$\displaystyle\left\{ \begin{array}{ll} & ye^{xy} = 3 \lambda x^2 \\ & xe^{xy} = 3 \lambda y^2 \\ \mathrm{constraint} & x^3 + y^3 = 16 \end{array} \right.$$ To solve this sytem, first notice that $x=y=0$ is not valid as it does not meet the constraint. So you can divide the first equation by the second to get the equation $\frac{y}{x} = \frac{y^2}{x^2}$, implying $y^3 = x^3$, or in other words, $x=y$. So you substitute these into the constraint equation and get $2x^3=16$ which has solution $x=2$. So $x=2=y$ is the only place where a maximum or minimum occurs, and in this case it is a maximum (why is it a maximum and not a minimum?).

1c.) You get $\nabla f = < 2xy^2z^2, 2x^2yz^2, 2x^2y^2z >$ and $\nabla g = < 2x, 2y, 2z >$. Therefore your system of equations is $$\left\{ \begin{array}{ll} & 2xy^2z^2 = 2 \lambda x \\ & 2x^2yz^2 = 2 \lambda y \\ & 2x^2y^2z = 2 \lambda z \\ \mathrm{constraint} & x^2+y^2+z^2=1 \end{array} \right.$$ One way to meet the first one is if $x=0$, but algebra in equations 2 and 3 shows that this will imply that $y=z=0$, but that's a problem as it does not satisfy the constraint (so $x \neq 0$). If $\lambda = 0$, then at most two of $x,y,$ and $z$ will equation zero (to meet the constraint), and such values will force $F(x,y,z)=0$.

Now assume that all three of $x,y,$ and $z$ are nonzero (this is the only choice we have left!). Solve each equation for $\lambda$ an set them equal to get the equation $\lambda = y^2z^2 = x^2z^2 = x^2y^2$. Now notice that you can write $F(x,y,z) = x^2y^2z^2 = \lambda x^2 = \lambda z^2 = \lambda y^2$.

So, $3x^2y^2z^2 = \lambda = y^2z^2$ implying that $3x^2=1$, so $x = \pm \frac{1}{\sqrt{3}}$. Similar arguments show that $y=z=\pm \frac{1}{\sqrt{3}}$. From these we get 8 possible choices, all of which give the value of $f$ to be $\frac{1}{27}$. Therefore the maximum value is $\frac{1}{27}$ and the minimum value is $0$ occurring at many many many places.

2.) Suppose your rectangle has sides of length $x$ and $y$. Then its area is $A(x,y) = xy$. We have a constraint of $x+y=p$ for some fixed constant real number $p$ (that is to say, $g(x,y) = x+y$). So $\nabla A = < y,x >$ and $\nabla g = <1,1 >$. This gives you the system $$\left\{ \begin{array}{ll} & y = \lambda \\ & x = \lambda \\ \mathrm{constraint} & x+y=p \end{array} \right.$$ It should be immediately clear that $y=x$ (they both equal $\lambda$!), so from the constraint equation we see that $2x = p$, or $x = \frac{p}{2}$. This forces the rectangle with optimal area to be a square with lengths $\frac{p}{2}$. Is this a maximum or a minimum? It must be a maximum as we can make a rectangle with such constraints have as little area as we want (how is that done?). (note: a rectangle with side length equal to zero is NOT a rectangle! So the minimum area isn't zero, but simply does not exist.)

3.) $\nabla P = < b \alpha L^{\alpha -1} K^{1 - \alpha}, b(1 - \alpha) L^{\alpha}K^{- \alpha} >$ and your constraint function is $g(L,K) = mL +nK$, so $\nabla g = < m, n >$. This generates the system of equations $$\left\{ \begin{array}{ll} & b \alpha L^{\alpha - 1} K^{1 - \alpha} = \lambda m \\ & b (1 - \alpha) L^{\alpha} K^{-\alpha} = \lambda n \\ \mathrm{constraint} & mL + nK = p \end{array} \right.$$ Rewrite the first equation as $b \alpha \left( \frac{L}{K} \right)^{\alpha - 1} = \lambda m$ and the second equation as $b(1-\alpha) \left( \frac{L}{K} \right)^{\alpha} = \lambda n$. Divide the first equation by the second and once the dust settles, you get $\frac{\alpha}{1- \alpha} \frac{K}{L} = \frac{m}{n}$. From this you can derive, say, $K = \frac{m}{m} L \frac{1-\alpha}{\alpha}$ and then plug this value into the constraint to get $L = \frac{p}{\alpha}{m}$. Use that value of $L$ to find the value of $K$.