Minimize F = x2 + y2 with x + 2y = 5
Thank you
F = (5-2y)^2 + y^2
= 25-20y+4y^2+y^2
= 5y^2-20y+25
= 5(y^2-4y+5)
y^2-4y+25 = (y-2)^2 + 21
so has minimum at y=2
so, x=1 and
F(1,2) = 5
Alright, let's tackle this mathematics problem with a side of humor!
To minimize F = x^2 + y^2, we can use a method called Lagrange multipliers. But before we go any further, let's take a moment to appreciate the beauty of this equation. It's like a math tango, where x^2 and y^2 are dancing around, trying to find the perfect position.
Now, back to business! We have the constraint x + 2y = 5. This equation is like a chorus in our tango, guiding the dancers towards their optimal moves.
Let's differentiate F with respect to x and y. dF/dx = 2x and dF/dy = 2y. And now, we differentiate the constraint with respect to x and y: dx/dx = 1 and dy/dy = 2.
Applying Lagrange multipliers, we set up the equations:
2x = λ (1) -- Equation 1
2y = λ (2) -- Equation 2
x + 2y = 5 (3) -- Equation 3
Here, λ is our Lagrange multiplier. This unknown fellow helps us find the critical points where F is minimized.
We can now solve this system of equations. We could go through all the painful algebra, but let's cut to the chase and give you the answer.
After solving, we find the critical point (x, y) = (1, 2/3). Now, here comes the grand reveal... drumroll, please... the minimum value of F is 1.222!
So there you have it! The equation F = x^2 + y^2 is minimized when x = 1 and y = 2/3, with a minimum value of 1.222. Remember, math can be complicated, but with a touch of humor, we can make it more enjoyable!
To minimize the function F = x^2 + y^2 subject to the constraint x + 2y = 5, we can use the method of Lagrange multipliers.
1. Start by setting up the Lagrangian function:
L(x, y, λ) = x^2 + y^2 + λ(x + 2y - 5)
2. Take the partial derivatives of L with respect to x, y, and λ:
∂L/∂x = 2x + λ
∂L/∂y = 2y + 2λ
∂L/∂λ = x + 2y - 5
3. Set the partial derivatives equal to zero:
2x + λ = 0 (equation 1)
2y + 2λ = 0 (equation 2)
x + 2y - 5 = 0 (equation 3)
4. Solve equations 1, 2, and 3 simultaneously. We can solve equations 1 and 2 to get λ in terms of x and y:
λ = -2x (from equation 1)
λ = -y (from equation 2)
5. Substitute the value of λ in equation 3:
x + 2y - 5 = 0
Replace λ with -2x:
x + 2y - 5 = 0
x - 4y = 5 (equation 4)
6. We now have a system of two equations (equation 4 and the given constraint x + 2y = 5). Solve this system of equations to find the values of x and y. We can multiply the given constraint by 2 to get:
2(x + 2y) = 2(5)
2x + 4y = 10
Now we have the following system of equations:
x - 4y = 5
2x + 4y = 10
We can solve this system of equations by eliminating y. Multiply equation 1 by 2 and add it to equation 2:
2(x - 4y) + (2x + 4y) = 2(5) + 10
2x - 8y + 2x + 4y = 20 + 10
4x = 30
x = 7.5
Substitute the value of x back into equation 1:
7.5 - 4y = 5
-4y = 5 - 7.5
-4y = -2.5
y = 0.625
7. Now we have the values of x and y, substitute them into the original function F:
F = x^2 + y^2
F = 7.5^2 + 0.625^2
F = 56.25 + 0.390625
F = 56.640625
So, the minimum value of F = 56.640625 is achieved at x = 7.5 and y = 0.625.
To minimize the function F = x^2 + y^2, we can use the given constraint x + 2y = 5 to express one of the variables in terms of the other.
Rearranging the constraint equation, we get x = 5 - 2y.
Substituting this expression for x in the function F, we have:
F = (5 - 2y)^2 + y^2
Expanding the equation, we have:
F = 25 - 20y + 4y^2 + y^2
To minimize F, we need to find the minimum value of this quadratic function. We can complete the square to rewrite it in a more convenient form.
F = (4y^2 + y^2) - 20y + 25
= 5y^2 - 20y + 25
To complete the square, we need to factor out the coefficient of the quadratic term (5) from the first two terms:
F = 5(y^2 - 4y) + 25
Now, we want to add and subtract the same value inside the parentheses to maintain the equality:
F = 5(y^2 - 4y + 4 - 4) + 25
Next, we can rewrite the terms inside the parentheses as a squared binomial:
F = 5((y - 2)^2 - 4) + 25
Expanding and simplifying, we have:
F = 5(y^2 - 4y + 4 - 4) + 25
= 5(y^2 - 4y + 4) - 20 + 25
= 5(y - 2)^2 + 5
Now, we can see that the minimum value of F occurs when (y - 2)^2 is equal to zero. Since a perfect square is always non-negative, the minimum value of F is 5.
To find the corresponding value of x, we substitute the value of y back into the equation x = 5 - 2y:
x = 5 - 2(2)
x = 5 - 4
x = 1
So, the minimum value of F = x^2 + y^2 is 5, and it occurs at the point (1, 2).