Question

Find the maximum value of u where u=sinx siny sin(x+y)

Answer 1

the max value of above question is 1/2

Answer 2

Step-by-step explanation:

Given: u=sinx siny sin(x+y)

To Find: Maximum value of u

Solution:

• Calculating partial derivatives of 'u' w.r.t. 'x' and 'y'

Since we have given, $u=sin(x) \ sin(y) \ sin(x+y)$, therefore its partial derivative with respect to $x$ is -

$\frac{\partial c}{\partial x} = \frac{\partial (sin(x) sin(y) sin(x+y))}{\partial x}$

$\Rightarrow \frac{\partial (sin(x) sin(y) sin(x+y))}{\partial x} = sin(y)[\frac{\partial sin(x)}{\partial x} sin(x+y) + \frac{\partial sin(x+y)}{\partial x} sin(x)]$

$\Rightarrow sin(y) [cos(x) sin(x+y) + cos(x+y) sin(x)] = sin(y) sin(2x+y)$

$\Rightarrow \frac{\partial u}{\partial x} = sin(y) sin(2x+y)$ . . . . . (1)

Similarly, $\Rightarrow \frac{\partial u}{\partial y} = sin(x) sin(2y+x)$ . . . . . . (2)

• Determining the values of x and y

At any stationary point, we know that $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = 0$. Therefore, putting (1) equal to zero, we get

$\Rightarrow \frac{\partial u}{\partial x} = sin(y) sin(2x+y) = 0$

Since $sin(y) \neq 0$ then $sin(2x+y) = 0 \Rightarrow 2x+y = n\pi$ . . . . . (3)

Similarly, $\Rightarrow \frac{\partial u}{\partial y} = sin(x) sin(2y+x) = 0 \Rightarrow 2y+x = n\pi$ . . . . . (4)

Equating (3) and (4), we get $x = y$ and substitute it in (3) to get $x = y = \frac{\pi}{3}$

• Calculating the maximum value of u at (π/3, π/3)

To find the maximum value of u at (π/3, π/3), considering,

$\Rightarrow \frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial}{\partial x}(\frac{\partial u}{\partial x}) = \frac{\partial (sin(y)sin(2y+x))}{\partial x} = 2sin(y)cos(2x+y)$ . . . . . . . (5)

Similarly,

$\Rightarrow \frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial}{\partial y}(\frac{\partial u}{\partial y}) = \frac{\partial (sin(x)sin(2y+x))}{\partial y} = 2sin(x)cos(2y+x)$ . . . . . . . (6)

And, $\Rightarrow \frac{\partial^{2} u}{\partial x \partial y}} = \frac{\partial}{\partial x}(\frac{\partial u}{\partial y}) = \frac{\partial (sin(x)sin(2y+x))}{\partial x} = sin(2y+2x) = \frac{\partial^{2} u}{\partial y \partial x}}$ . . . . . . (7)

at the point $(\frac{\pi}{3}, \frac{\pi}{3})$

$\frac{\partial^{2} u}{\partial x^{2}} = 2sin(\frac{\pi}{3})cos(\frac{2\pi}{3}+\frac{\pi}{3}) = -\sqrt{3}$  . . . . (8)

$\frac{\partial^{2} u}{\partial y^{2}} = 2sin(\frac{\pi}{3})cos(\frac{2\pi}{3}+\frac{\pi}{3}) = -\sqrt{3}$ . . . . (9)

and, $\Rightarrow \frac{\partial^{2} u}{\partial x \partial y}} = sin(\frac{2\pi}{3} +\frac{2\pi}{3}) = \frac{-\sqrt{3}}{2}$ . . . . (10)

Substituting (8) (9) and (10) in $(\frac{\partial^{2} u }{\partial x^{2} })(\frac{\partial^{2} u }{\partial y^{2} }) - (\frac{\partial^{2} u }{\partial x \partial y })^{2}$ you will get

$(\frac{\partial^{2} u }{\partial x^{2} })(\frac{\partial^{2} u }{\partial y^{2} }) - (\frac{\partial^{2} u }{\partial x \partial y })^{2} = (-\sqrt{3})(-\sqrt{3}) - (\frac{\sqrt{3}}{2})^{2} = \frac{9}{4} &> 0$ . . . . . (11)

Equation (11) implies, at the points $(\frac{\pi}{3}, \frac{\pi}{3})$, we will get the maximum value of u. Therefore, $u=sin(\frac{\pi}{3}) \ sin(\frac{\pi}{3}) \ sin(\frac{\pi}{3} + \frac{\pi}{3}) = \frac{3\sqrt{3}}{8}$

Hence, the maximum value of $u = \frac{3\sqrt{3}}{8}$ at the points $(\frac{\pi}{3}, \frac{\pi}{3})$