SOLUTION: 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)?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 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)?      Log On


   

Question 1150766: 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)?
Found 3 solutions by ikleyn, MathLover1, Edwin McCravy:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

The difference is

    p(x+y)-q(x+y) = 2x^2 + 3y^2 - 4x + 18y + 39


Continue by completing the squares

    = (2x^2 - 4x)      + (3y^2 + 18y) + 39

    = 2*(x^2 - 2x)     + 3*(y^2 + 6y) + 39 =

    = 2*(x^2 - 2x + 1) + 3*(y^2 + 6y + 9) + 39 - 2 - 27 =

    = 2*(x-1)^2        + 3*(y+3)^2        + 10.


The minimum is achieved at x= 1,  y= -3  and is equal to  10.    ANSWER

Solved.


Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

p%28x%2By%29+=+2x%5E2+%2B3y%5E2
q%28x%2By%29=4x-18y-39
p%28x%2By%29-q%28x%2By%29=2x%5E2+%2B3y%5E2-%284x-18y-39%29
p%28x%2By%29-q%28x%2By%29=2x%5E2+%2B3y%5E2-4x%2B18y%2B39

2x%5E2+%2B3y%5E2-4x%2B18y%2B39=0
%282x%5E2+-4x%29%2B%283y%5E2%2B18y%29%2B39=0
2%28x%5E2+-2x%2Bb%5E2%29-2b%5E2%2B3%28y%5E2%2B6y%2Bb%5E2%29-3b%5E2%2B39=0
2%28x%5E2+-2x%2B1%5E2%29-2%2A1%5E2%2B3%28y%5E2%2B6y%2B3%5E2%29-3%2A3%5E2%2B39=0
2%28x+-1%29%5E2-2%2B3%28y%2B3%29%5E2-27%2B39=0
2%28x+-1%29%5E2%2B3%28y%2B3%29%5E2-29%2B39=0
2%28x+-1%29%5E2%2B3%28y%2B3%29%5E2%2B10=0

so, minimum value will be %28x+-1%29%5E2=0=>x=1 and
%28y%2B3%29%5E2=0=>x=-3

minimum value will be when+p%28x%2By%29-q%28x%2By%29=0%2B0%2B10=10


Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
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