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Q.
If sum of two numbers is 24 and their product is as large as possible, then one of the number is
Application of Derivatives
Solution:
Let one number be $x$. Then, the other number is $(24-x)$.
( $\because$ sum of the two numbers is 24$)$ Let $y$ denotes the product of the two numbers. Thus, we have
$y=x(24-x)=24 x-x^2$
On differentiating twice w.r.t. $x$, we get
$\frac{d y}{d x}=24-2 x \text { and } \frac{d^2 y}{d x^2}=-2$
Now, put $ \frac{d y}{d x}=0$
$\Rightarrow 24-2 x =0$
$\Rightarrow x =12$
At $x=12, \frac{d^2 y}{d x^2}=-2 < 0$
By second derivative test, $x=12$ is the point of local maxima of $y$. Thus, the product of the numbers is maximum when the numbers are $12$ and $24-12=12$.
Hence, the numbers are $12$ and $12$.