How to maximize this expression?

Question

Find the maximum value of $$\sqrt{x} + \sqrt{y} + \sqrt{z}$$ under the constraint $$2x + 3y + 5z = 15.$$

Can this problem be solved using Lagrange multipliers? If yes, how?

Answer 1

$1)$ Let $F(x,y,z,\lambda) \equiv F= \sqrt x+\sqrt y+\sqrt z + \lambda(2x+3y+5z-15)$

$2)$ Compute $F_x = F_y = F_z = F_\lambda = 0$

$3)$ Find $x,y,z$ in terms of $\lambda$

$4)$ Use $x,y,z$ (in terms of $\lambda$) in $2x+3y+5z =15$ and find the value(s) of $\lambda$ and hence those of $x,y,z$.

$5)$ From the set(s) of values of $x,y,z$ find which one corresponds to the maximum value of $\sqrt x+\sqrt y+\sqrt z$

Answer 2

I will give you the actual solution only so that you may verify your answer at the end. If this is a homework problem, you would be committing some pretty egregious academic misconduct if you just took this answer and ran with it.

Spoiler: The solution is $x=\frac{225}{62}, y=\frac{50}{31}, z=\frac{18}{31}$.

If you had a copy of Wolfram Mathematica you could have generated this answer using the code Maximize[{x^(1/2) + y^(1/2) + z^(1/2), 15 == 2 x + 3 y + 5 z}, {x, y, z}]

As for how you might approach the question properly, we start by taking the Lagrangean $$\mathcal{L} = \sqrt{x} + \sqrt{y} + \sqrt{z} + \lambda(15 - 2x -3y -5z)$$

Then we get the first order conditions

\begin{align} 0= \frac{\partial \mathcal{L}}{\partial x} &= \frac{1}{2\sqrt{x}} - 2\lambda &\implies \lambda = \frac{1}{4\sqrt{x}} \\ 0= \frac{\partial \mathcal{L}}{\partial y} &= \frac{1}{2\sqrt{y}} - 3\lambda &\implies \lambda = \frac{1}{6\sqrt{y}}\\ 0= \frac{\partial \mathcal{L}}{\partial z} &= \frac{1}{2\sqrt{z}} - 5\lambda &\implies \lambda = \frac{1}{10\sqrt{z}}\\ 0= \frac{\partial \mathcal{L}}{\partial \lambda} &=15-2x-3y-5z &\implies 2x+3y+5z=15 \end{align}

Can you think of how to solve these?

Answer 3

I think by C-S is better: $$15=\frac{(2x+3y+5z)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}\geq\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}=\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\frac{31}{30}},$$ which gives $$\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\sqrt{15.5}.$$ The equality occurs for $$\left(\sqrt{2x},\sqrt{3y},\sqrt{5z}\right)||\left(\frac{1}{\sqrt2},\frac{1}{\sqrt3},\frac{1}{\sqrt5}\right),$$ which says that we got a maximal value.