Please advise on how to solve this exercise
A firm’s profit is given by the function: $$\text{Profit} = 1200 - 6x^2 - 8x + 4xy - 10y^2 + 96y$$ where $x$ denotes output and $y$ advertising expenditure. Find the level of output and advertising expenditure to maximalize profit. Find the level of profit for this point. Confirm that second order conditions are satisfied. Find the level of maximum profit.
As $x,y$ goes to infinity, profit goes to negative infinity. So, the maximum exists. We need to check the critical values. As economics only cares about positive numbers, critical values are $x=0,y=0$ and the values that make the partial derivatives $0$.
For $x=0$, the profit is $$p(y)=1200-10y^2+96y$$ $p'(y)=96-20y$, thus, $p$ will be maximum at $y_0=24/5$. Plugging this into $p$, profit will be $\frac{7152}5=1430.4$
For $y=0$, the profit is $$q(y)=1200-6x^2-8x$$ which is definitely decreasing for $x\ge0$. So, its maximum will be at $x_0=0$. Plugging this into $q$, the profit will be $1200$.
Let $$f(x,y)=1200-6x^2-8x+4xy-10y^2+96y$$ Then, $$\frac{\partial f}{\partial x}=-12x-8+4y\qquad\frac{\partial f}{\partial y}=-20y+96+4x$$ In order to optimize $f$ we need to make the partial derivatives zero. Let $(x_0,y_0)$ be the tuple that makes $f$ maximal. We need to solve $$4y-12x=8\implies y-3x=2\qquad (1)$$ $$20y-4x=96\implies 5y-x=24\qquad (2)$$ So, $$(2)-5(1)\implies14=24-5\cdot2=(5y_0-2x_0)-5(y_0-3x_0)=5y_0-x_0-5y_0+15x_0=14x_0$$ Thus, $x_0=1$. Plugging this into (1), we have $$y_0=3x_0+2=5$$ For the second order condition, $$\frac{\partial^2 f}{\partial x^2}=-12\qquad\frac{\partial^2 f}{\partial y\partial x}=4\qquad \frac{\partial^2 f}{\partial y^2}=-20$$
So, the Hessian matrix is $$\begin{pmatrix} -12&4\\ 4&-20 \end{pmatrix}$$ , which is clearly negative-definite; because, top-left entry is negative and the determinant $(-12)\cdot(-20)-4\cdot4=240-16=224>0$. So, the critical value, indeed gives a local maximum. So, $(1,5)$ maximizes the profit. Plugging these two values in gives $$f(x,y)=1436$$
Among the three possible profits, the maximum is obtained for $(x_0,y_0)=(1,5)$, which is $1436$.
