Question

if X,Y,Z are angles of a triangle then find the function u=sinX.sinY.sinZ is

Answer 1

Answer:

thank for free points

Step-by-step explanation:

Answer 2

Answer:

The angles of the triangle are all equal to 60° and the maximum function value is $3\sqrt{3} /}{8}$.

Step-by-step explanation:

Given $x,y,z$ are angles of a triangle.

Sum of the angles in a triangle is 180°. Thus

$x+y+z=180^{o}$

And let the function be written as

$u(x,y,z)=sin(x)sin(y)sin(z)$

Let us take the Lagrange multiplier to generate the system of equations that gives the value of critical points.

$\implies \lambda(x+y+z)=\nabla\big(sin(x)sin(y)sin(z)\big)$

where $\nabla$ represents the gradient operator.

Writing the partial derivatives of the equation, we get

$\lambda=cos(x)sin(y)sin(z)$                                        ...(1)

$\\\ \lambda=sin(x)cos(y)sin(z)$                                       ...(2)

$\lambda=sin(x)sin(y)cos(z)$                                       ...(3)

All the three equations are equal to λ. So, let us consider first two equations and equate them

$cos(x)sin(y)sin(z)=sin(x)cos(y)sin(z)$

$\implies cos(x)sin(y)=sin(x)cos(y)$

$\implies \dfrac{sin(x)}{cos(x)} =\dfrac{sin(y)}{cos(y)}$

$\implies tan(x) =tan(y)\implies x =y$                        ...(4)

Now consider equation (2) & (3) and equate them.

$cos(x)sin(y)sin(z)=sin(x)sin(y)cos(z)$

$\implies cos(x)sin(y)=sin(x)cos(z)$

$\implies \dfrac{sin(x)}{cos(x)} =\dfrac{sin(z)}{cos(z)}$

$\implies tan(x) =tan(z)\implies x =z$                          ...(5)

Comparing equations (4) & (5), we get $x=y=z$.

Thus, from $x+y+z=180^{o}$

$\implies x+x+x=180^{o}$

$\implies3x=180^{o}\implies x=\dfrac{180}{3} =60^{o}$

Hence, the critical point of the functions is

$(60,60,60)$

and the function is

$u(60,60,60)=sin(60)sin(60)sin(60)$

                   $=\bigg(\dfrac{\sqrt{3} }{2}\bigg )^{3} =\dfrac{3\sqrt{3} }{8}$

Therefore, the angles of the triangle are all equal to 60° and the maximum function value is $3\sqrt{3} /}{8}$.

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