Calculus/Extreme points
Expert: Paul Klarreich - 2/6/2011
Questionhi, i need help with this question.
The function given by f(x; y) = xy cos(xy) has a critical point at (0; 0).
Is it a local minimum, local maximum or a saddle point?
A heated metal plate P is represented by a rectangle in the xy-plane, with vertices (0; 0), (4; 0), (4; 2) and (0; 2). At each point (x; y) of P the temperature, in degrees Celsius, is given by
T(x; y) = 10x2 - 20xy + 20y + 20
Find the point(s) at which the temperature of the plate is greatest and find the greatest temperature.
thank you.
AnswerQuestioner: julie
Country: Australia
Category: Calculus
Private: No
Subject: calculus.
Question: hi, i need help with this question.
The function given by f(x; y) = xy cos(xy) has a critical point at (0; 0).
Is it a local minimum, local maximum or a saddle point?
A heated metal plate P is represented by a rectangle in the xy-plane, with
vertices (0; 0), (4; 0), (4; 2) and (0; 2). At each point (x; y) of P the
temperature, in degrees Celsius, is given by
T(x; y) = 10x2 - 20xy + 20y + 20
Find the point(s) at which the temperature of the plate is greatest and find
the greatest temperature.
thank you.
......................................
Here are my computations for the first:
fx = y (cos(xy) - xy sin(xy))
fxx = y(-y sin(xy) - y( sin (xy) + xy cos (xy)))
= -2 y^2 sin(xy) - xy^3 cos (xy)
fy = x (cos(xy) - xy sin(xy))
fyy = -2 x^2 sin(xy) - x^3y cos (xy)
fxy = cos(xy) - xy sin(xy) + y(- x sin(xy) - x sin(xy) - x^2y cos(xy))
fxy = cos(xy) - xy sin(xy) - y(2x sin(xy) + x^2y cos(xy))
Now you want to compute D = fxx fyy - (fxy)^2 at (0,0)
See:
http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/pd9.html
fxx = y(- y sin(xy) - y^2 cos(xy)) = 0
fyy = 0
fxy = 1
D = - 1
D < 0 --> neither local max or min
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To check these, try:
http://www.calculator-grapher.com/derivative-calculator.html
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To do your second (which I won't):
1. Find your extreme points WITHIN the region.
2. Compute T at each of them.
3. Compute T at each of the CORNERS.
Pick the biggest.