5
$\begingroup$

I'm trying to solve the following problem:

Production of a certain company depends on $3$ inputs $x,y,z$ in the following way:

$$f(x,y,z) = 50x^{\frac{2}{5}} y^{\frac{1}{5}} z^{\frac{1}{5}}$$

Budget of the company is $24 000$ dollars and products $x, y, z$ can be bought for $80$, $12$ or $10$ dollars per unit in that order. What combination of inputs results in biggest production?

It's a Calculus test preparation problem and I have no idea how I would start. Could you help me?

My idea is to use constrained extremas but I don't know how.

$\endgroup$
  • $\begingroup$ If you can buy quantities in of any size, i.e. $x,y,z \in \mathbb{R}$, then write $x = (24000-12y-10z)/80$. Now solve it by maximising $f$ as a function of $y, z$ by finding the partial derivatives. $\endgroup$ – fGDu94 Jan 26 at 17:12
  • $\begingroup$ I did the partial derivation: $\frac{\partial f}{y}: 10x^{\frac{2}{5}}z^{\frac{1}{5}y^{\frac{4}{5}}}$ $\frac{\partial f}{\partial z}: 10x^{\frac{2}{5}}y^{\frac{1}{5}}z^{\frac{-4}{5}}$ . I don't know what to do now. $\endgroup$ – Emanuel Jan 26 at 17:34
  • $\begingroup$ You should substitute $x=(24000-12y-10z)/80$ first then differentiate in $y,z$ $\endgroup$ – fGDu94 Jan 26 at 17:35
  • $\begingroup$ @fGDu94 thank you very much! $\endgroup$ – Emanuel Jan 26 at 18:14
4
$\begingroup$

Let $g(x,y,z)=80x+12y+10z$ and use Lagrange multipliers: \begin{align} 20 x^{-3/5} y^{1/5} z^{1/5} &= 80 \lambda\\ 10 x^{2/5} y^{-4/5} z^{1/5} &= 12 \lambda\\ 10 x^{2/5} y^{1/5} z^{-4/5} &= 10 \lambda\\ 80x+12y+10z &= 24000 \end{align} The resulting solution is $(x,y,z)=(150,500,600)$.

$\endgroup$
  • $\begingroup$ how did you get the $x,y,z$ in the end? $\endgroup$ – Emanuel Jan 26 at 17:44
  • 1
    $\begingroup$ One way is to multiply the first equation by $x$, the second equation by $2y$, and the third equation by $2z$ to obtain $80\lambda x=24\lambda y=20\lambda z$. Now divide by $\lambda$ and substitute $y$ and $z$ in the fourth equation, which you can then solve for $x$. $\endgroup$ – Rob Pratt Jan 26 at 17:50
7
$\begingroup$

By AM-GM $$24000=80x+12y+10z=2(20x+20x+6y+5z)\geq8\sqrt[4]{20^2\cdot6\cdot5x^2yz}.$$ The equality occurs for $20x=6y=5z$ or $(x,y,z)=(150,500,600)$.

I.e., $x^2yz$ (and so $50(x^2yz)^{\frac{1}{5}}$) attains a maximal value for $$(x,y,z)=(150,500,600).$$

$\endgroup$
  • 1
    $\begingroup$ +1 I always like it when the simplest tools can be used. I was fooled by the tags into rushing into calculus! $\endgroup$ – almagest Jan 27 at 6:27
4
$\begingroup$

This is a standard Lagrange multiplier question. We want to maximize $$50x^{2/5}y^{1/5}z^{1/5}$$ subject to $80x+12y+10z=24000$.

So we try to maximise $$50x^{2/5}y^{1/5}z^{1/5}-\lambda(80x+12y+10z)$$ We know the maximum must be on the boundary or at a stationary point. But $xyz=0$ on the boundary, so it will be at a stationary point. Setting the three partial derivatives to 0 gives us three equations relating $x,y,z$ from which we easily deduce $y=10x/3,z=4x$. Substituting into the constraint then gives $$x=150,y=500,z=600$$

$\endgroup$
1
$\begingroup$

You are trying to maximize $f(x, y, z) = 50x^{\frac{2}{5}} y^{\frac{1}{5}} z^{\frac{1}{5}}$ subject to $g(x,y,z) = 80x + 12y + 10z = 24000$. To make it simpler, where the maximum of $f$ is is the same as where the maximum of $h(x, y, z) = x^2yz$ is. You can use the method of Lagrange multipliers to solve this.

We want to find $x, y, z$ such that $$\nabla h = \lambda\nabla g$$ where $\nabla h$ is the gradient of $h$. This then gives us the three equations $$2xyz = 80\lambda$$ $$x^2z = 12\lambda$$ $$x^2y = 10\lambda$$

From the constraint function, we also have that $$80x + 12y + 10z = 24000$$

Now that there are $4$ equations and $4$ variables, it is possible to solve for $x, y, z$, and $\lambda$. Although there are solutions where one of $x, y, z$ is $0$, these are extraneous because production would be $0$. The actual answer is then $$x = 150, y = 500, z = 600$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.