Posted by **summer** on Sunday, February 12, 2012 at 8:21pm.

Q=3x^2+4y^2 if x+y=7 what is minimum value for Q.

I have absolutely no idea how to solve this problem.

- math -
**Reiny**, Sunday, February 12, 2012 at 10:10pm
from the 2nd equation: y = 7-x

sub into the other

Q = 3x^2 + 4(7-x)^2)

= 3x^2 + 4(49 - 14x + x^2

= -x^2 + 196 - 56x + 4x^2

= 3x^2 - 56x + 196

The minimum of Q occurs at the vertex of the corresponding parabola

If you know Calculus ...

dQ/dx = 6x - 56 = 0 for a min of Q

6x=56

x = 56/6 = 28/3

Q = 3(28/3)^2 - 56(28/3) + 196 = -196/3

If no Calculus, then complete the square ...

Q = 3(x^2 - (56/3)x + 784/9 - 784/9) + 196

= 3( (x - 28/3)^2 - 784/9) + 196

= 3(x-28/3)^2 - 784/3 + 196

= 3(x-28/3)^2 - 196/3

so the min of -196/3 happens when x = 28/3

- math -
**grave**, Tuesday, December 10, 2013 at 9:13pm
What is the distance between the parallel lines, whose equation are 3x + 4y = 2 and 3x + 4y = -5.

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