I consider this as a 3D problem in Calc 3.
The second partials test:
If (a,b) is a point on f(x,y) for which both partial derivatives are 0, then
find the quantity
d = fxx(a,b)∙fyy(a,b)-[fxy(a,b)]2
1. If d is negative, then (a,b) is a saddle point which is neither a relative
maximum nor minimum.
2. If d is positive
a. if fxx(a,b) > 0, then (a,b) is a relative minimum
b. if fxx(a,b) < 0, then (a,b) is a relative maximum
3. If d is 0, the test fails.
if p(x+y) = 2x²+3y², and q(x+y)=4x-18y-39, where x and y are real numbers, what is the minimum value of p(x+y)-q(x+y)?
f(x,y) = p(x+y)-q(x+y) = (2x^2 +3y^2)-(4x-18y-39) = 2x²+3y²-4x+18y+39,
which is an elliptical paraboloid whose axis of symmetry is parallel to the
z-axis, and has just one minimum point or one maximum point.
We set the two partial derivatives of f equal to 0:
fx=px-q=4x-4, fy=py-qy=6y+18
4x-4=0, 6y+18=0
x=1, y=-3
So if that's the minimum point, that's the one we use to evaluate the minimum
value. But we ought to show that it's a minimum
We calculate d
d = fxx(1,-3)∙fyy(1,-3)-[fxy(1,-3)]2
fxx(1,-3) = 4, fyy(1,-3) = 6, ffxx(1,-3) = 0,
d = 4∙6-0² = 24 > 0}
So since d = 24 > 0 and fxx(1,-3) = 4 then f does have a relative minimum
at (1,-3).
That minimum value is found by substituting
f(1,-3) = p(1+(-3))-q(1+(-3)) = 2(1)²+3(-3)²-4(1)+18(-3)+39,
That works out to be 10.
Answer: 10
Edwin