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.
What is the distance between the parallel lines, whose equation are 3x + 4y = 2 and 3x + 4y = -5.
To find the minimum value of Q, we can use the method of completing the square. Here's how you can solve the problem step-by-step:
Step 1: Rearrange the equation x + y = 7 to solve for y:
y = 7 - x
Step 2: Substitute the value of y into the original equation for Q:
Q = 3x^2 + 4(7 - x)^2
Step 3: Expand the equation:
Q = 3x^2 + 4(49 - 14x + x^2)
= 3x^2 + 196 - 56x + 4x^2
= 7x^2 - 56x + 196
Step 4: Let's express the quadratic function Q in vertex form by completing the square:
Q = 7(x^2 - 8x) + 196
= 7(x^2 - 8x + 16) - 112 + 196
= 7(x - 4)^2 + 84
Step 5: Since the square of a real number is always non-negative, the lowest possible value of (x - 4)^2 is 0. Therefore, the minimum value of Q occurs when (x - 4)^2 = 0. So, x - 4 must equal 0, solving for x:
x - 4 = 0
x = 4
Step 6: Substituting the value of x back into the equation y = 7 - x, we can find the value of y:
y = 7 - 4
y = 3
Step 7: Now we can find the minimum value of Q by substituting the values of x and y back into the original equation for Q:
Q = 3(4)^2 + 4(3)^2
= 48 + 36
= 84
Therefore, the minimum value of Q is 84 when x = 4 and y = 3.
To find the minimum value of Q = 3x^2 + 4y^2 when x + y = 7, we can use the method of differentiation. Here's how you can solve it step by step:
Step 1: Start by expressing one variable in terms of the other using the given equation x + y = 7. Let's solve for x:
x = 7 - y
Step 2: Substitute the value of x in the equation for Q:
Q = 3(7 - y)^2 + 4y^2
Step 3: Expand and simplify the equation:
Q = 3(49 - 14y + y^2) + 4y^2
Q = 147 - 42y + 3y^2 + 4y^2
Q = 147 - 42y + 7y^2
Step 4: Take the derivative of Q with respect to y. This will help us find the critical points:
Q' = -42 + 14y
Step 5: Set the derivative equal to zero and solve for y to find the critical points:
-42 + 14y = 0
14y = 42
y = 3
Step 6: Substitute the value of y back into the equation x + y = 7 to find the value of x:
x + 3 = 7
x = 7 - 3
x = 4
Step 7: Now we have the critical point (x, y) = (4, 3). To determine whether it is a minimum or maximum, we'll take the second derivative:
Q'' = 14
Since the second derivative is positive (Q'' > 0), it indicates that the critical point is a minimum.
Step 8: Finally, substitute the values of x and y into the equation for Q to find the minimum value:
Q = 3(4)^2 + 4(3)^2
Q = 3(16) + 4(9)
Q = 48 + 36
Q = 84
Therefore, the minimum value of Q is 84.
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