The values of x and y are 70/22 and 30/22 respectively by using the first-order condition of differential calculus.
<h3>What is the first-order condition in differential calculus?</h3>
A first-order differential equation is represented by the equation
with 2 variables x & y, including its function f(x,y) specified on a xy-plane.
Given that:
![\mathbf{f(x,y) =-22x^2+22xy-11y^2+110x-40y-23}](https://tex.z-dn.net/?f=%5Cmathbf%7Bf%28x%2Cy%29%20%3D-22x%5E2%2B22xy-11y%5E2%2B110x-40y-23%7D)
Let us first differentiate the above equation with respect to x, we have:
![\mathbf{\dfrac{\partial f(x,y) }{\partial x} = -44x +22y -0+110-0-0=0}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Cpartial%20f%28x%2Cy%29%20%7D%7B%5Cpartial%20x%7D%20%20%3D%20-44x%20%2B22y%20-0%2B110-0-0%3D0%7D)
(multiply by -1)
44x - 22y = 110 ------ (equation 1)
Now, differentiating with respect to y, we have:
![\mathbf{\dfrac{\partial f(x,y) }{\partial y} =0 +22x-22y +0-40-0=0}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Cpartial%20f%28x%2Cy%29%20%7D%7B%5Cpartial%20y%7D%20%20%3D0%20%2B22x-22y%20%2B0-40-0%3D0%7D)
![\mathbf{\implies 22x-22y -40=0}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cimplies%2022x-22y%20-40%3D0%7D)
22x - 22y = 40 ----- (equation 2)
Now, we have a system of equations:
44x - 22y = 110
- ---- ( subtracting equation 2 from 1; elimination method)
<u> 22x - 22y = 40 </u>
<u>22x + 0 = 70 </u>
<u />
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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