Minimize Q = 3x^2
+ 4y^2, where x + y = 7.

The min value of Q is Q = 84 and it happens when x = 4 and y = 3

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.

the square root of what number? if you add that, then someone will probably answer the question.

range is infinite on both x and y axis positive and negative

step-by-step explanation: