# Find the maximum and minimum values attained by the function f along the path c(t).

$f(x,y)= xy; \space c(t) = (\cos(t), \sin(t)); \space 0 \leq t \leq 2 \pi$

$f(x,y) = x^2 + y^2; \space c(t)= (\cos(t), 8 \sin(t)); \space 0 \leq t \leq 2 \pi$

This problem refers to calculus and aims to understand that over a closed and bounded interval, the continuous function of one variable always reaches the maximum and minimum values. The weights of the range of the function are always finite.

In this problem, we are given a function and path that the function is being estimated along. We have to calculate the maximum and minimum associated with the function along the path.

Part a:

Given that, $f(x,y)= xy$ and $c(t) = (\cos(t), \sin(t));$ for $0 \leq t \leq 2 \pi$.

$f(x,y)= xy$

$f(x,y)= \cos(t). \sin(t)$

Using the trigonometric formula $\sin(2x)= 2 \sin(x)\cos (x)$:

$\sin(x) \cos(x)$ is equals to $\dfrac{\sin(2x)}{2}$.

Inserting $\sin(x) \cos(x)$ in $f(x,y)$:

$f(x,y)= \dfrac{\sin(2x)}{2}$

We know that the range of sine function is always between $-1$ to $1$, that is:

$-1 \leq \sin(2x) \leq 1$

$\dfrac{-1}{2} \leq \dfrac{ \sin(2x)}{2} \leq \dfrac{1}{2}$

$\dfrac{-1}{2} \leq f(x,y) \leq \dfrac{1}{2}$

Part b:

Given that $f(x,y)= x^2+y^2$ and $c(t) = ( \cos(t), 8\sin (t));$ for $0 \leq t \leq 2 \pi$.

$f(x,y)= x^2 + y^2$

$f(x,y)= (\cos (t))^2. (8 \sin (t))^2$

$f(x,y)= \cos^2 t. 64 \sin^2 t$

Using the trigonometric formula $\sin^2(x) + \cos^2(x) = 1$,

$\cos^2(t)$ is equals to $1 – \sin^2(t)$.

Inserting the new $\cos^2(t)$ in $f(x,y)$:

$f(x,y)= 1 -\sin^2(t) + 64 \sin^2(t)$

$f(x,y)= 1 + 63 \sin^2(t)$

We know that the range of $\sin^2 (t)$ function is always between $0$ to $1$, that is:

$0 \leq \sin^2(t) \leq 1$

$0 \leq 63 \sin^2(t) \leq 63$

$1 \leq 1+ 63\sin^2(t) \leq 64$

$1 \leq f(x,y) \leq 64$

Part a: Maximum and minimum value attained by the function $f(x,y) = xy$ along the path $(cos(t), sin(t))$ is $\dfrac{-1}{2}$ and $\dfrac{1}{2}$.

Part b: Maximum and minimum value attained by the function $f(x,y = x^2 + y^2)$ along the path $( \cos(t), 8\sin (t))$ is $1$ and $64$.

## Example

Find the maximum and minimum range of the function $f$ along the path $c(t)$

$-(b) \space f(x,y) = x^2 + y^2; \space c(t)= (\cos(t), 4 \sin(t)); \space 0 \leq t \leq 2 \pi$

Given, $f(x,y)= x^2+y^2$ and $c(t) = ( \cos(t), 4\sin (t));$ for $0 \leq t \leq 2 \pi$.

$f(x,y)= x^2+y^2$

$f(x,y)= \cos^2 t. 16 \sin^2 t$

Using the trigonometric formula $\sin^2(x)+ \cos^2(x)=1$,

$\cos^2 (t)$ is equals to $1 – \sin^2 (t)$.

$f(x,y)$ becomes:

$f(x,y)=1 -\sin^2(t)+16 \sin^2 (t)$

$f(x,y)=1+15 \sin^2(t)$

Range of $\sin^2 (t)$ function is between $0$ to $1$, that is:

$0 \leq \sin^2(t) \leq 1$

$0 \leq 15 \sin^2(t) \leq 15$

$1 \leq 1+ 15\sin^2(t) \leq 16$

$1 \leq f(x,y) \leq 16$