Set up the Lagrangian function:
$$\min_{k,l} rk+wl-\lambda\left((k^{\alpha}l^{1-\alpha})^{1/\beta}-y\right)$$
Note that the function $(k^{\alpha}l^{1-\alpha})^{1/\beta}$ is quasi-concave and homogeneous and of degree less than 1 (since $\beta>1$). A quasi-concave function of degree between 0 and 1 is concave. The negative of a concave function is convex ($\lambda$ is positive). Further, the two linear functions $rk$ and $wl$ are convex. The sum of convex functions is convex. Thus, the entire Lagrangian function is convex. The first order conditions will be sufficient for a minimum. It is also noted that the production function has infinite marginal product for either capital or labor when one is set to $0$. Thus, we will not have to worry about corner solutions where only one of the two inputs is used. Further, prices are strictly positive and so no more than y
will be produced. The first order conditions are:
$$r=\lambda\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\alpha\left(\frac{l}{k}\right)^{1-\alpha}$$
$$w=\lambda\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}$$
$$(k^{\alpha}l^{1-\alpha})^{1/\beta}=y$$
Solving these gives:
$$\frac{r}{\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\alpha\left(\frac{l}{k}\right)^{1-\alpha}}=\frac{w}{\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}}$$
$$\frac{\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}}{\alpha\left(\frac{l}{k}\right)^{1-\alpha}}=\frac{w}{r}$$
Gives one key relationship:
$$\frac{k}{l}=\frac{\alpha}{1-\alpha}\frac{w}{r}$$
Now, we use this along with the production constraint to solve for the inputs:
$$(k^{\alpha}l^{1-\alpha})=y^{\beta}$$
$$\left(\frac{\alpha}{1-\alpha}\frac{w}{r}\right)^{\alpha}l=y^{\beta}$$
$$l^{*}=\left(\frac{1-\alpha}{\alpha}\frac{r}{w}\right)^{\alpha}y^{\beta}$$
$$k^{*}=\left(\frac{\alpha}{1-\alpha}\frac{w}{r}\right)^{1-\alpha}y^{\beta}$$
Plugging these back into the cost function gives the desired result:
$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(\frac{\alpha}{1-\alpha}\right)^{1-\alpha}+\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$
$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(1+\left(\frac{\alpha}{1-\alpha}\right)\right)\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$
$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(\frac{1}{1-\alpha}\right)\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$
$$y^{\beta}w^{1-\alpha}r^{\alpha}\frac{1}{\alpha^{\alpha}\left(1-\alpha\right)^{1-\alpha}}$$
By the envelope theorem, the derivative of an optimized function with respect to one of its parameters can be taken treating the choice variables as fixed. At the optimum, a very small change in these choice variables has no effect on the optimized value. This means, when we take the derivative of the optimized cost function with respect to $r$ and $w$ we will get back $k^*$ and $l^*$ respectively. I will leave this to you to check.
Since there are two quantities you can set independently this is a two-variable calculus problem. You should call the quantity produced by the first process $q_1$ and the second quantity $q_2$ and then write an expression $P(q_1,q_2)$ for the profit. To find the local extrema of the profit, you take the partial derivative with respect to each parameter and set both of them equal to zero. This will give you two equations in two variables to solve.
It seems a bit odd that it would be a multivariate calculus problem given the prerequisites and background, but that's my best interpretation of the problem.
Best Answer
a) Look up what it means to be concave/convex. Sketch a graph of C. Based on the graph, does it look like C is concave/convex? How might you prove it?
b) What is the definition of "average cost"? Now this is a standard calc 1 exercise: Find the global minimum of the average cost.