Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I took the partial derivatives of $\sin(x)+\sin(y)+\sin(z)$ and it didn't work out, so I am trying to use Lagrange's method (with the constraint: $x+y+z=\pi$)... I am not sure how to set this up.

EDIT: while I appreciate other approaches, if someone could direct me towards using the Lagrange's method it would help me learn how to use that method as well :)

share|cite|improve this question
up vote 5 down vote accepted

Using Lagrange multipliers, and setting up the initial function:

$$\Lambda(x, y, z, \lambda)=\sin(x)+\sin(y)+\sin(z)+\lambda(x+y+z-\pi)$$

Setting the gradient $\nabla\Lambda=0$, we can write:

$$\frac{\partial\Lambda}{\partial x}=\cos(x)+\lambda=0$$ $$\frac{\partial\Lambda}{\partial y}=\cos(y)+\lambda=0$$ $$\frac{\partial\Lambda}{\partial z}=\cos(z)+\lambda=0$$ $$\frac{\partial\Lambda}{\partial \lambda}=x+y+z-\pi=0$$

The original equations give us $x=\arccos(-\lambda), y=\arccos(-\lambda),z=\arccos(-\lambda)$, substituting into the final partial derivative gives us:


We can then use this to find the critical points, which are when $x=y=z=\frac{\pi}{3}$.

Plugging this value into our initial equation gives us the maximum value: $3\sin(\frac{\pi}{3})=\frac{3\sqrt{3}}{2}$.

In order to find the minimum, we note that any of the infinite number of $\cos^{-1}(-\lambda)$ are critical points, providing $x+y+x=\pi$. As we are looking to now minimize our solution, it makes sense to maximize one solution and minimize two others.

We can do this by observing that $\cos^{-1}(-\frac{1}{2})=2\pi-\frac{\pi}{3}$, and $\cos^{-1}(-\frac{1}{2})=-\frac{\pi}{3}$, plugging these values into our equation gives us:


Minimizing our objective function.

share|cite|improve this answer

Note that $\sin(x)$ is a concave function. For any concave function $f(x)$, we have that $$\dfrac{\displaystyle \sum_{k=1}^{n} f(x_k)}{n} \leq f \left( \dfrac{\sum_{k=1}^{n} x_k}{n} \right)$$

One intuitive way to think about this is that the center of mass of a $n$ body system lying on a concave curve always lies below the curve. Taking $f(x)$ to be $\sin(x)$ and $n=3$, we get that $$\dfrac{\sin(x) + \sin(y) + \sin(z)}{3} \leq \sin \left( \dfrac{x+y+z}{3} \right) = \sin(\pi/3) = \dfrac{\sqrt{3}}{2}$$ Hence, $$\sin(x) + \sin(y) + \sin(z) \leq \dfrac{3 \sqrt{3}}{2} $$ Note that the equality holds when $x = y= z = \pi/3$.

Below is a figure for the explanation.enter image description here I have drawn the curve $y=sin(x)$ but you can draw this for any concave curve $y=f(x)$.

The image was made using grapher.

share|cite|improve this answer
beautiful way of explanation ! – Theorem Jun 3 '12 at 18:08

For another approach, the constraint gives us $z = \pi - x -y$, so you can solve the problem by finding the minimum and maximum value of the function $$f(x,y) = \sin(x) + \sin(y) + \sin(\pi - x -y).$$

share|cite|improve this answer
I actually attempted this originally and took the partial derivatives, but didn't seem to get anywhere. would you perhaps elaborate? – Donnie Jun 3 '12 at 22:02

Probably you aren't looking at the trivial answer...

i,j,k integer numbers (including zero and negatives)

max(( f(x,y,z) ) = 3

min(( f(x,y,z) ) = -3

Xmax(i) = pi * (1/2 + 2*i)

Ymax(j) = pi * (1/2 + 2*j)

Zmax(k) = pi * (1/2 + 2*k)

Xmin(i) = pi * (-1/2 + 2*i)

Ymin(j) = pi * (-1/2 + 2*j)

Zmin(k) = pi * (-1/2 + 2*k)

Easy answer because of f(x,y,z) = g(x) + h(y) + q(z)

Where g,h,q are "limited" (-1 <= g,h,q <= +1).

Max(f(x,y,x) = Max(g(x)) + Max(h(y)) + Max(q(z))

Min(f(x,y,z) = Min(g(x)) + Min(h(y)) + Min(q(z))

Max(g(x)) = g(Xmax(i)) = 1

Max(h(y)) = h(Xmax(j)) = 1

Max(q(z)) = q(Xmax(k)) = 1

Min(g(x)) = g(Xmin(i)) = -1

Min(h(y)) = h(Xmin(j)) = -1

Min(q(z)) = q(Xmin(k)) = -1

share|cite|improve this answer
This doesn't really answer the question since $x+y+z=\pi$. Besides, the question is more than a year old and already has several answers. It might be a better idea to find new questions and contribute by answering them. Welcome to math.stackexchange! – mrf Sep 3 '13 at 20:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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