A pair of numbers that have a sum of 50
Let the number is x, so the other number is 50 - x
Let f(x) be the largest product so:
![f(x)=\text{ x(50-x)}](https://img.qammunity.org/2023/formulas/mathematics/college/1v3itdii2isbtg2iysdvetdk0htxlras9z.png)
Simplify the expression :
![\begin{gathered} f(x)=\text{ x(50-x)} \\ f(x)=50x-x^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3kx1jgnmp780gfzymbtge08j78gxmc7uva.png)
Diffrentiate with respect to x
![\begin{gathered} f(x)=\text{ x(50-x)} \\ f(x)=50x-x^2 \\ \text{ Diffrentiate with respect to x} \\ f^(\prime)(x)=50-2x \\ \text{Apply derivative equal to zero:} \\ 50-2x=0 \\ 50=2x \\ x=25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mdrojtieiiqj7c7uirmuaeb6cmfp256zmr.png)
Now for to check for the f(x) is maximum for x = 25
Calculate the second derivative and put x = 25 is the f(x) is negative then the multiplication f(x) is maximum
![\begin{gathered} f^(\prime)(x)=50-2x \\ \text{ Differentiate with respect to x} \\ f^(\prime)^(\prime)(x)=0-2 \\ \text{ Substitute x = 25} \\ f^(\doubleprime)(25)=-2 \\ f^(\doubleprime)(25)<0 \\ \text{Thus the function f(x) is maximum for x = 25} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/73nnkf55125usdjozbnmglq4hvdfaqq9e5.png)
Thus, the first number is 25
Second number is : 50 -x = 50-25 = 25
Numbers are 25, 25
Answer : 25 + 25 =50 (sum)
25 * 25 = 625 (maximum possible product)