Which one is true for the extrema of f(x, y)=x^2-4x+4/3 y^3-4y The graph of f(x,

Question

Which one is true for the extrema of f(x, y)=x^2-4x+4/3 y^3-4y The graph of f(x, y) has a relative maximum at the point(2,1,-20/3) The graph of f(x, y) has a relative maximum at the point(2,1,-20/3) The graph of f(x, y) has a relative maximum at the point(2,1,-20/3) The Second Partials Test fails None of these Evaluate the double integral: integral_e^2^e^3 integral_1^4 2y/x dy dx 3 15 1/3 15 ln 3/2 Which is the correct procedure to calculate the double integral: integral_0^1 integral_1^x ln(y) dy dx integral_0^1 dx.integral_1^x ln y dy integral_0^1(x ln x-x) dx integral_0^1 (x ln x-x-1) dx integral_0^1(x ln x-x+1) dx None of these

Explanation / Answer

7)f(x,y)=x2 -4x +(4/3)y3 -4y

for critical points fx =0 , fy =0

fx =2x-4=0 , fy =4y2-4=0

x=4/2,y2=4/4

x=2,y =1 ,y =-1

(x,y)=(2,1),(2,-1) are critical points

f(2,1)=4-8+(4/3)-4 =-20/3

f(2,-1)=4-8-(4/3)+4=-4/3

fxx=2, fyy =8y ,fxy =0

D=fxx fyy -fxy2

at (2,1)

fxx=2, fyy =8 ,fxy =0

D=16>0

relative minimum at (2,1,-20/3)

at (2,-1)

fxx=2, fyy =-8 ,fxy =0

D=-16<0

relative maximum at (2,-1,-4/3)

the graph of f(x,y) has relative minimum at (2,1,-20/3)

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