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.
You have made a typo at the exercise. $\text{It is "A builder has purchased 21,00}$ $\color{red}{0}$ $\text{square meters of land}$ $\text{on which it is planned to build... "}$
But you have written it at your constraint: $600X+300Y \leq 21,000 \ \ $ (Square meters Constraint)
If you use this constraint the problem is feasible. This is what I got by using the Excel solver:
Excel sheet with formulas
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Best Answer
You minimize the cost function. The cost function is $3K+6L$, since the prices are respectively $3$ euros and $6$. Then the output is $600$. Now you set this value equal to the output function to obtain the constraint.
$$25\cdot K^{1/3}\cdot L^{2/3}=600 \ \ \Rightarrow \ \ 600-25\cdot K^{1/3}\cdot L^{2/3}=0$$
Indeed you now apply the method of Lagrange multipliers. the Lagrangian function is
$$\mathcal L=3K+6L+\lambda\left(600-25\cdot K^{1/3}\cdot L^{2/3}\right)$$
I think you can proceed. Give a reply if it works for you.