Question 1150766
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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 = f<sub>xx</sub>(a,b)∙f<sub>yy</sub>(a,b)-[f<sub>xy</sub>(a,b)]<sup>2</sup>

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 f<sub>xx</sub>(a,b) > 0, then (a,b) is a relative minimum
   b. if f<sub>xx</sub>(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:

f<sub>x</sub>=p<sub>x</sub>-q=4x-4, f<sub>y</sub>=p<sub>y</sub>-q<sub>y</sub>=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 = f<sub>xx</sub>(1,-3)∙f<sub>yy</sub>(1,-3)-[f<sub>xy</sub>(1,-3)]<sup>2</sup>

f<sub>xx</sub>(1,-3) = 4, f<sub>yy</sub>(1,-3) = 6, ff<sub>xx</sub>(1,-3) = 0, 

d = 4∙6-0² = 24 > 0}

So since d = 24 > 0 and f<sub>xx(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</pre>