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
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.
2 answers
What is the distance between the parallel lines, whose equation are 3x + 4y = 2 and 3x + 4y = -5.