Maximize a function Resolve the following problem. Let $\alpha, \beta > 0, p_{1},p_{2} > 0$ fixed. Maximize $x^{\alpha}y^{\beta}$ subject to $p_{1}x + p_{2}y = 10$ $x,y > 0$.
 A: This is a geometric programming problem, a type of convex optimization problem. There are efficient solvers available.
However, if you are looking for an analytical solution for this particular variant, then we use a log transform and substitution to get:
$$x=\frac{10-p_2y}{p_1} \implies z:= \ln\left(x^{\alpha}y^{\beta}\right)= \alpha \ln(x) + \beta\ln(y) = \alpha\ln\left(\frac{10-p_2y}{p_1}\right) + \beta\ln(y)$$
We can further simplify this by pulling out $p_1$:
$$z:= \alpha\ln\left(10-p_2y\right) - \alpha\ln(p_1) + \beta\ln(y)$$
We can now apply the usual calculus to get the stationary point:
$$\text{Find }y:\frac{dz}{dy} = \frac{\beta}{y}-\frac{\alpha p_2}{10-p_2y}=0$$
Setting both sides equal to each other we get:
$$\frac{\beta}{y}=\frac{\alpha p_2}{10-p_2y} \implies \beta\left(10-p_2y\right)=\alpha p_2 y \implies 10\beta=(\alpha+\beta)p_2y \implies y=\frac{10\beta}{(\alpha+\beta)p_2}$$
Since $\alpha, \beta, p_2 > 0$ we know that $y>0$
Taking the second derivative of $z$ wrt $y$, we get:
$$\frac{d^2z}{dy^2} = -\frac{\alpha p_2^2}{(10 - p_2y)^2} - \frac{\beta}{y^2}$$
Plugging in our optimal $y$ (i.e, $y^*$) and simplifying we get:
$$\left.\frac{d^2z}{dy^2}\right|_{y=y^*}=-\frac{p_2^2}{100}\left(\frac{\alpha }{\left(1- \frac{\beta}{\alpha+\beta}\right)^2} + \frac{\beta }{\left(\frac{\beta}{\alpha+\beta}\right)^2}\right) < 0 \;\;\;\forall \alpha,\beta > 0$$
Therefore, $y^*$ is a occurs at the maximum. Since $y^*$ is unique, it is the global maximum.
From our substitution, we get:
$$x=\frac{10-p_2y}{p_1} \implies x = \frac{10-p_2\frac{10\beta}{(\alpha+\beta)p_2}}{p_1} = \frac{10-\frac{10\beta}{(\alpha+\beta)}}{p_1} = 10\left(\frac{1-\frac{\beta}{(\alpha+\beta)}}{p_1}\right)$$
Again, since $\alpha,\beta>0$ we will have $0<\frac{\beta}{(\alpha+\beta)} < 1 \implies x>0$.
Since $\ln(w)$ is monotonic increasing in $w$,
$$\arg\limits_{x,y} \max z = \arg\limits_{x,y} \max e^z = x^{\alpha}y^{\beta}$$
Therefore, the values $x^*,y^*$ that optimize the original problem are given by:
$$y=\frac{10\beta}{(\alpha+\beta)p_2}, \;\;\;x= 10\left(\frac{1-\frac{\beta}{(\alpha+\beta)}}{p_1}\right)$$
We have shown that $\alpha,\beta,p_1,p_2> 0 \implies x^*,y^*>0$
