# Math Help - maximize with Lagrange

1. ## maximize with Lagrange

Hey any idea and help is much appreciated Many thanks b

2. Hello, Bobby!

4. An advertising agency spends $x$ dollars on a newspaper campaign
and a further $y$ dollars promoting client's products on local radio.
It receives a 15% commission on all sales that the client receives.
The agency has \$10,000 to spend in total.

The client earns $M$ dollars from its sales
where: . $M \;=\;\frac{100,000x}{50 + x} +\frac{40000y}{30 + y}$

Use the method of Lagrange multipliers to determine how much should be spent
on advertising in newspapers and on radio to maximize the agency's net income.
We have the function: . $M \;=\;10^5\!\cdot\!\frac{x}{x+50} + 4\!\cdot\!10^4\!\cdot\!\frac{y}{y+30}$
and the constraint: . $x + y \:\leq \:10^4$

Our function is: . $F\;=\;10^5\!\cdot\!\frac{x}{x+50} + 4\!\cdot\!10^4\!\cdot\!\frac{y}{y+30} - \lambda(x + y - 10^4)$

Set the three partial derivatives equal to zero.

. . $\frac{\partial F}{\partial x} \: = \: \frac{50\!\cdot\!10^5}{(x+50)^2} - \lambda \: = \: 0\quad{\color{blue}[1]}$

. . $\frac{\partial F}{\partial y} \: = \:\frac{12\!\cdot\!10^5}{(y+30)^2} - \lambda \: = \: 0\quad{\color{blue}[2]}$

. . $\frac{\partial F}{\partial\lambda} \: = \;\;\; x + y - 10^4 \;\; = \: 0 \quad{\color{blue}[3]}$

From ${\color{blue}[1]}$ and ${\color{blue}[2]}$, we have: . $\frac{50\!\cdot\!10^5}{(x+50)^2} \;=\;\lambda \;=\;\frac{12\!\cdot\!10^5}{(y+30)^2}$

. . Then: . $(y+30)^2 \;=\;\frac{6}{25}(x+50)^2\quad\Rightarrow\quad y \;=\;\frac{\sqrt{6}}{5}(x + 50) - 30$

Substitute into ${\color{blue}[3]}$: . $x + \left[\frac{\sqrt{6}}{5}(x + 50) -30\right] \;=\;10^4\quad\Rightarrow\quad x + \frac{\sqrt{6}}{5}x + 10\sqrt{6} - 30 \;=\;10,000$

. . $\left(\frac{5+\sqrt{6}}{5}\right)x \;=\;10(1003-\sqrt{6})\quad\Rightarrow\quad x \;=\;\frac{50(1003-\sqrt{6})}{5 + \sqrt{6}} \;=\;6715.564051$

Therefore: . $\begin{Bmatrix}x & = & \6715.56 \\ y & = & \3284.44\end{Bmatrix}$

Note: the agency's income is $0.15M$

And someone check my work . . . please!
.