math
posted by summer on .
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.

from the 2nd equation: y = 7x
sub into the other
Q = 3x^2 + 4(7x)^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(x28/3)^2  784/3 + 196
= 3(x28/3)^2  196/3
so the min of 196/3 happens when x = 28/3 
What is the distance between the parallel lines, whose equation are 3x + 4y = 2 and 3x + 4y = 5.