# What combination of inputs results in the largest output?

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.

• 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. – fGDu94 Jan 26 at 17:12
• 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. – Emanuel Jan 26 at 17:34
• You should substitute $x=(24000-12y-10z)/80$ first then differentiate in $y,z$ – fGDu94 Jan 26 at 17:35
• @fGDu94 thank you very much! – Emanuel Jan 26 at 18:14

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 $$(x,y,z)=(150,500,600)$$.

• how did you get the $x,y,z$ in the end? – Emanuel Jan 26 at 17:44
• 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 $x$. – Rob Pratt Jan 26 at 17:50

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

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

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 $$x=150,y=500,z=600$$

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 $$x = 150, y = 500, z = 600$$