Respuesta :
The values of x and y are 70/22 and 30/22 respectively by using the first-order condition of differential calculus.
What is the first-order condition in differential calculus?
A first-order differential equation is represented by the equation [tex]\mathbf{ \dfrac{dy}{dx} =f (x,y) }[/tex]with 2 variables x & y, including its function f(x,y) specified on a xy-plane.
Given that:
[tex]\mathbf{f(x,y) =-22x^2+22xy-11y^2+110x-40y-23}[/tex]
Let us first differentiate the above equation with respect to x, we have:
[tex]\mathbf{\dfrac{\partial f(x,y) }{\partial x} = -44x +22y -0+110-0-0=0}[/tex]
[tex]\mathbf{\implies -44x +22y+110=0}[/tex] (multiply by -1)
44x - 22y = 110 ------ (equation 1)
Now, differentiating with respect to y, we have:
[tex]\mathbf{\dfrac{\partial f(x,y) }{\partial y} =0 +22x-22y +0-40-0=0}[/tex]
[tex]\mathbf{\implies 22x-22y -40=0}[/tex]
22x - 22y = 40 ----- (equation 2)
Now, we have a system of equations:
44x - 22y = 110
- ---- ( subtracting equation 2 from 1; elimination method)
22x - 22y = 40
22x + 0 = 70
x = 70/22
Replacing the value of x into equation (1), we have:
44x - 22y = 110
44(70/22) - 22y = 110
140 - 22y = 110
140 - 110 = 22y
30 = 22y
y = 30/22
Learn more about the first-order conditions in differential calculus here;
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