Answer:
![x1=√(216) \ and \ y1=\ √(216)](https://img.qammunity.org/2021/formulas/mathematics/college/ybm5dureh4qdbl8y5uozlk5cyyrf5v81na.png)
Explanation:
Let the first number is x1 and other number is y1 then
![x1 * y1 =216](https://img.qammunity.org/2021/formulas/mathematics/college/5umb3opicyo096clk3rvhfyq9uqh4330mt.png)
Therefore
![y1=](https://img.qammunity.org/2021/formulas/mathematics/college/nn7uuibqlz5x9a4umuz1n4jokz566jv50o.png)
![(216)/(x1)](https://img.qammunity.org/2021/formulas/mathematics/college/mkc3vrdx8zmvrkrgpa4k9uiew07dinewiy.png)
also there sum is
....Eq(1)
Putting the value of y1 in the previous equation
........Eq(2)
Differentiate the the Eq(2) with respect to x1 we get
![(ds1)/(dx1) \ =\ 1+216*(1)/(-x1^(2) )](https://img.qammunity.org/2021/formulas/mathematics/college/z7nwtrn53s08jkr5vkdrcwd80tzygufym4.png)
![(ds1)/(dx1) \ =\ 1-(216)/(x1^(2) )\ =\ 0](https://img.qammunity.org/2021/formulas/mathematics/college/ij9duk99l72te2c7sclfydkcgm7gf6z0o3.png)
![{x1^(2) }\ =\ 216\\ x1=√(216)](https://img.qammunity.org/2021/formulas/mathematics/college/60zlvj6897hopyb213hq6celgyvc05jor0.png)
Putting the value of X1 in Eq(1) we get
![y1=(216)/(√(216) ) \\y1=(216*√(216) )/(216) \\y1=\ √(216)](https://img.qammunity.org/2021/formulas/mathematics/college/7fdt3xkr6vl8o07q58sjyjzhwr7bllkza4.png)
So
![x1=√(216) \ and \ y1=\ √(216)](https://img.qammunity.org/2021/formulas/mathematics/college/ybm5dureh4qdbl8y5uozlk5cyyrf5v81na.png)