Answer to Question #107932 in Calculus for annie

Question

Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤225.

1)What is the maximum value of f(x,y) ?

2) List the point(s) where the function attains its maximum as an ordered pair, such as (-6,3), or a list of ordered pairs if there is more than one point, such as (1,3), (-4,7).

3) What is the minimum value of f(x,y)?

4) List points where the function attains its minimum as an ordered pair, such as (-6,3), or a list of ordered pairs if there is more than one point, such as (1,3), (-4,7).

Answer 1

ANSWER: 1)${ f }_{ max }=674$ at the points $(-2,\sqrt { 221 } )$ and $(-2,-\sqrt { 221 } )$

2) ${ f }_{ min }=-7$ at the point $(1,0)$

EXPLANATION

Since the function $f(x,y)=2{ x }^{ 2 }+3{ y }^{ 2 }-4x-5$ is continuous on the domain $C=\left\{ (x,y):x^{ 2 }+y^{ 2 }\le 225 \right\}$ (compact set) the function takes the largest and smallest values. If ${ \ f }_{ max }\ (or\ { \ f }_{ min })=f(a,b)\ and\ (a,b)\in \left\{ (x,y):x^{ 2 }+y^{ 2 }<225 \right\} ,\\ then\quad \nabla f(a,b\ )=0$

It is necessary to compare the value of the functional such points with the values of the function at the boundary points of the region $C$ .

Step 1)

$$\nabla f(x,y\ )=\left( { f }_{ x }^{ ' },f_{ y }^{ ' } \right) =0\quad \Leftrightarrow \begin{cases} 4x-4=0 \\ 6y=0 \end{cases}\quad \Leftrightarrow \quad \quad \begin{cases} x=1 \\ y=0 \end{cases}\quad .$$

$(1,0)\in \ \left\{ (x,y):{ x }^{ 2 }+{ y }^{ 2 }<225 \right\} ,\ f(1,0)=2\cdot 1-4\cdot 1-5=-7$

Step 2)

At the boundary of the region $C$ function $f$ has the form $f(x,\ \pm \sqrt { 225-{ x }^{ 2 } } )=450+(225-{ x }^{ 2 })-4x-5=$ $=-{ x }^{ 2 }-4x+670,\quad x\in [-15,15].$ ${ \left( -{ x }^{ 2 }-4x+670 \right) }_{ x }^{ ' }=-2x-4.=0,\quad if\quad x=-2.\quad$Therefore, we compare the

values of the function $φ(x)=-{ x }^{ 2 }-4x+670$ at the points $x=-15,\ x=-2,\ x=15$ . Or for function $f$ at the points $(-15,0)\ ,\ (-2,\pm \sqrt { 225-4 } )=(-2,\pm \sqrt { 221 } \ ),\ (15,0)\quad$

Step 3)

Compare the value of the function $f$ at the points $(1,0)\ , (-15,0)\ , (-2,-\sqrt { 221 } )\ ,(-2,\sqrt { 221 } )\ , (15,0)\quad$

$f(1,0)=-7,f(-15,0)=f(15,0)=385,\quad f(-2,\sqrt { 221 } ) =$

$=f(-2,-\ \sqrt { 221 } \ )=674\quad$

So, ${ f }_{ max }=674$ at the points $(-2,\sqrt { 221 } )\quad and\quad (-2,-\sqrt { 221 } )$ .${ f }_{ min }=-7\quad$at the point $(1,0)$