11.Find all of the critical points of ... and use the Seconds Partials Test to c
ID: 2887468 • Letter: 1
Question
11.Find all of the critical points of ... and use the Seconds Partials Test to classify them as relative maxima, relative minima, or saddle points.
7. Draw the contour plots for the following functions b.) f(x y) (x - y)2 8. For the function f(x.y) -x3- 3xy +y2 a.) Find the directional derivative of f at the point (1,2) if we start to move towards the point (3,5). b.) In what direction does f experience the greatest rate of increase at the point (1,2)? What is the greatest speed in f at the same point? 9. For the given surface z ye a.) Find the equation of the tangent plane to the surface at the point (3,1,4) 10. As a solid right circular cylinder is heated, its radius r and height h increase; hence, so does its surface area S. Suppose that at the instant when r 10 cm and h 100 cm, r is increasing at 0.2 cm/br and h is increasing at 0.5 cm/hr. How fast is S increasing at this instant? 11. Find all of the critical points of f(x,y)-x2 + 3y3-7xy + 5 and use the Seconds Partials Test to classify them as relative maxima, relative minima, or saddle points. 12. Set up, but do not evaluate, an integral that gives the mass of the solid in the first octant bounded by the cylinder ?? + z 2-1 and the planes x-1 and x-4 if the density at each point is given by f(x,y, z)-z?Explanation / Answer
Given f(x,y) = x^2 + 3y^3 - 7xy + 5
fx = d/dx ( x^2 + 3y^3 - 7xy + 5)
= 2x - 7y
fxx = 2
fxy = - 7
fy = d/dy ( x^2 + 3y^3 - 7xy + 5)
= 0 + 3( 3y^2 ) - 7x
= 9y^2 - 7x
fyy = 2( 9y) => 18y
fyx = - 7
for critical points
fx = 0 and fy = 0
2x - 7 y = 0 and 9y^2 - 7x = 0
2x = 7y
x = (7/2) y
9y^2 - 7( 7/2) y =0
9y^2 - 49/2 y = 0
y ( 9y - 49/2) = 0
y = 0 , 9y - 49/2 = 0
y = 0 , 9y = 49/2 => y = 49/18
at y = 0 at y = 49/18
2x = 7y 2x = 7( 49/18)
2x = 7(0) x = 7(49/36)
x = 0 x = 343/36
points are ( 0,0) and ( 49/18, 343/36 )
Next, we use the second derivative test to classify them.
D(a,b) = fxx(a,b) fyy(a,b) - ( fxy(a,b) * fyx(a,b) )
or
= fxx(a,b) fyy(a,b) - ( fxy(a,b))^2
If D(a,b) > 0 and fxx(a,b) < 0, then f(x,y) has a
maximum at (a,b)
If D(a,b) > 0 and fxx(a,b) > 0, then f(x,y) has a
minimum at (a,b)
If D(a,b) < 0 and , then f(x,y) has neither a maximum nor a
minimum at (a,b) i,e Saddle Point.
fxx(x,y) = 2 , fyy = 18 y , fyx = fxy = - 7
at (0.0)
D(0,0) = fxx(0,0) * fyy(0,0) - [ fxy(0,0) ]^2
= 2 * 18(0) - [ -7 ]^2
= 0 - 49
= -49
D(0,0) < 0 (0,0) is saddle point
at ( 49/18 ,343/36)
D( 49/18 ,343/36) = fxx( 49/18 ,343/36) * fyy( 49/18 ,343/36) - [ fxy( 49/18 ,343/36) ]^2
= 2 * 18( 343/36) - [ -7 ]^2
= 343 - 49
= 294
D( 49/18 ,343/36) > 0 , fxx( 49/18 ,343/36) > 0 then f(x,y) has a
minimum at ( 49/18 ,343/36)
f( 49/18 ,343/36) = (343/36)^2 + 3( 49/18)^2 - 7(343/36) ( 49/18) + 5
= -25.25952
Minimum value is f( 49/18 ,343/36) = -25.25952
There is no Maximum
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