Answer: The min value of Q is Q = 84 and it happens when x = 4 and y = 3
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Explanation:
x+y = 7 turns into y = 7-x after subtracting x from both sides
Replace y with 7-x in the other equation to get
Q = 3x^2 + 4y^2
Q = 3x^2 + 4( y )^2
Q = 3x^2 + 4(7-x)^2
Q = 3x^2 + 4(49 - 14x + x^2)
Q = 3x^2 + 196 - 56x + 4x^2
Q = 7x^2 - 56x + 196
We have a function with one variable. Graphing 7x^2-56x+196 produces a parabola in which the vertex point is what we're after
Anything in the form p(x) = ax^2+bx+c will have a vertex (h,k) such that
h = -b/(2a)
k = p(h)
Let's find the x coordinate of the vertex
h = -(-56)/(2*7)
h = 4
Use this to find the y coordinate of the vertex
k = p(h)
p(x) = 7x^2-56x+196
p(h) = 7h^2-56h+196
p(4) = 7(4)^2-56(4)+196
p(4) = 84
The vertex is the lowest point in this case (since a = 7 is positive) and the vertex is (4,84)
Therefore, the minimum value of Q is Q = 84 and this happens when x = 4 and y = 3. Recall that y = 7-x.
We can see that,
Q = 3x^2 + 4y^2
Q = 3(4)^2 + 4(3)^2
Q = 3(16) + 4(9)
Q = 48 + 36
Q = 84
Which helps us verify we have the right Q value.
