(1 point) find the maximum and minimum values of the function f(x, y)=2x2+3y2−4x−5 on the domain x2+y2≤289.the maximum value of f(x, y) is: 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).the minimum value of f(x, y) is: 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).

the maximum value inside the region x2+y2≤289 is f( 0 , √289 )= 862 and the minimum is f(1,0) = -7

Step-by-step explanation:

f(x,y)=2x2+3y2−4x−5 , x2+y2≤289

we can divide the calculation in 2 sections: inside the boundaries and in the boundaries of the region

inside the boundaries)

x2+y2=289

denoting fb(x,y) as the function evaluated in the boundaries:

fb(x,y)=2x2+3y2−4x−5 = 2(x2+y2) + y2−4x−5 = 2*289 + (289 - x2) - 4x - 5 =

fb(x,y)= 3*289 - x2 - 4x -5,  which clearly decreases with an increase of x and viceversa. Thus:

(xmax)2+(0)*2=289 → xmax=√289 → fb min =  3*289 - (√289)2 - 4(√289) -5 = 2*289-4√289 -5 = 505

(xmin)2+289=289 → xmin=0, → fb max =  3*289 - (0)2 - 4(0) -5 = 3*289 -5 = 862

inside the boundaries

f(x,y)=2x2+3y2−4x−5

-partial derivative with respect with x) 4x-4 = 0 → x=1

- partial derivative with respect with y) 6y= 0 → y=0

f(1,0) = 2(1)2 +3(0)2 -4(1) -5 = 2 - 4 -5 = -7

therefore

f(1,0) = -7

f( √289 , 0)=  505

f( 0 , √289 )= 862

since f is a continuous function in the defined region

the maximum value is  f( 0 , √289 )= 862 and the minimum is f(1,0) = -7

alot

step-by-step explanation:

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