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How do I solve the following problem. I'm not sure what it falls under so I don't know where to look in the book for help....

Find the three positive number with product 27 which minimizes the sum of the third with twice the second with four times the first.

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up vote 1 down vote accepted

Hint: So we want to minimize $4x+2y+z$ subject to the condition that $x$, $y$. and $z$ are positive and $xyz=27$.

If you have already seen Lagrange Multipliers, this is a good place to use the method. If you have not, see the alternate approach below. Let $$F(x,y,z,\lambda)=4x+2y+z+\lambda(xyz-27),$$ and set the partial derivatives of $F$ with respect to $x$, $y$, $z$, and $\lambda$ equal to $0$. Solve and see what this tells you. You will need to check that there really is a minimum at the point you find.

Another way: We have $z=\frac{27}{xy}$. So we want to minimize $G(x,y)$, where $$G(x,y)=4x+2y+\frac{27}{xy}.$$ Calculate the partial derivatives of $x$ and $y$ with respect to $x$ and $y$, and set them equal to $0$. Solve. Then you have to show that we really do get a minimum.

Remark: You are doing this for a calculus course, so I have suggested calculus ways. But there is (in my opinion) a nicer way. Let $u=4x$, $v=2y$, and $w=z$. We want to minimize $u+v+w$ subject to the condition that the variables are positive, and $uvw=(27)(8)$.

By the Arithmetic Mean Geometric Mean Inequality (known affectionately as AM-GM to generations of Math Olympiad contestants) we have $$\frac{u+v+w}{3}\ge (uvw)^{1/3},$$ with equality only when $u=v=w$.

So $\frac{u+v+w}{3}\ge 6$, with equality when $u=v=w$. That gives minimum value of $u+v+w$ equal to $18$. Putting $u=v=w$ we find that each is equal to $6$, so our minimum is reached at $x=\frac{3}{2}$, $y=3$, and $z=6$.

The above remark is intended to show that calculus is not the only tool for such problems. It will also serve as a check for your Lagrange Multiplier (or other) calculation.

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This is a constrained optimization problem. Let's call the 3 numbers $x,y,z$. Your objective is to minimize:

$$ f(x,y,z)=4x+2y+z $$

subject to the constraint:

$$ g(x,y,z)=xyz=27 $$

To solve the problem, use Lagrange multipliers:

$$ \mathcal{L}(x,y,z,\lambda)=f(x,y,z)-\lambda(g(x,y,z)-27) $$ So we require $\nabla\mathcal{L}=0$:

$$ 4=\lambda yz\\ 2=\lambda xz\\ 1=\lambda xy\\ xyz=27 $$

You can solve for the unknowns $x,y,z$. If you obtain multiple answers, choose the positive ones.

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