Final answer:
To minimize Q=5x^2+2y^2 with the constraint x+y=7, we express y in terms of x, substitute into Q, and find the vertex of the parabola. The minimum value occurs at x=2 and y=5, satisfying the constraint.
Step-by-step explanation:
To minimize the function Q=5x^2+2y^2 with the constraint x+y=7, we can express one variable in terms of the other using the given equation. Let's solve for y first:
y = 7 - x
Now substitute this expression for y into the function Q:
Q = 5x^2 + 2(7-x)^2
Expand and simplify the equation:
Q = 5x^2 + 2(49 - 14x + x^2)
Q = 5x^2 + 98 - 28x + 2x^2
Q = 7x^2 - 28x + 98
Now let's find the vertex of the parabola, which gives the minimum value of Q, by completing the square or using the vertex formula. The x-coordinate of the vertex is given by -b/(2a), where a = 7 and b = -28:
x = -(-28)/(2*7) = 28/14 = 2
Now we substitute x = 2 back into the equation x+y=7 to find y:
y = 7 - 2 = 5
Thus, the values that minimize the function Q are x = 2 and y = 5.
To check if the answer is reasonable, we can verify that these values of x and y indeed satisfy the original constraint x+y=7:
2 + 5 = 7, which is true.