# maximize with Lagrange

• Jul 30th 2007, 04:59 AM
bobby87
maximize with Lagrange
Hey any idea and help is much appreciated Many thanks b
• Jul 30th 2007, 06:11 AM
Soroban
Hello, Bobby!

Quote:

4. An advertising agency spends $\displaystyle x$ dollars on a newspaper campaign
and a further $\displaystyle y$ dollars promoting client's products on local radio.
It receives a 15% commission on all sales that the client receives.
The agency has $10,000 to spend in total. The client earns$\displaystyle M$dollars from its sales where: .$\displaystyle M \;=\;\frac{100,000x}{50 + x} +\frac{40000y}{30 + y}$Use the method of Lagrange multipliers to determine how much should be spent on advertising in newspapers and on radio to maximize the agency's net income. Give your answers correct to two decimal places. We have the function: .$\displaystyle M \;=\;10^5\!\cdot\!\frac{x}{x+50} + 4\!\cdot\!10^4\!\cdot\!\frac{y}{y+30}$and the constraint: .$\displaystyle x + y \:\leq \:10^4$Our function is: .$\displaystyle F\;=\;10^5\!\cdot\!\frac{x}{x+50} + 4\!\cdot\!10^4\!\cdot\!\frac{y}{y+30} - \lambda(x + y - 10^4)$Set the three partial derivatives equal to zero. . .$\displaystyle \frac{\partial F}{\partial x} \: = \: \frac{50\!\cdot\!10^5}{(x+50)^2} - \lambda \: = \: 0\quad{\color{blue}[1]}$. .$\displaystyle \frac{\partial F}{\partial y} \: = \:\frac{12\!\cdot\!10^5}{(y+30)^2} - \lambda \: = \: 0\quad{\color{blue}[2]}$. .$\displaystyle \frac{\partial F}{\partial\lambda} \: = \;\;\; x + y - 10^4 \;\; = \: 0 \quad{\color{blue}[3]}$From$\displaystyle {\color{blue}[1]}$and$\displaystyle {\color{blue}[2]}$, we have: .$\displaystyle \frac{50\!\cdot\!10^5}{(x+50)^2} \;=\;\lambda \;=\;\frac{12\!\cdot\!10^5}{(y+30)^2} $. . Then: .$\displaystyle (y+30)^2 \;=\;\frac{6}{25}(x+50)^2\quad\Rightarrow\quad y \;=\;\frac{\sqrt{6}}{5}(x + 50) - 30$Substitute into$\displaystyle {\color{blue}[3]}$: .$\displaystyle 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$. .$\displaystyle \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$Therefore: .$\displaystyle \begin{Bmatrix}x & = & \$6715.56 \\ y & = & \$3284.44\end{Bmatrix}$Note: the agency's income is$\displaystyle 0.15M\$

And someone check my work . . . please!
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