Question
Find two numbers whose sum is 24 and whose product is as large as possible.

Answer

Let the two numbers be x and y.
According to the question, $\text{x}+\text{y}=24\ \Rightarrow\ \text{y}=24-\text{x}\ \dots\text{(i)}$
And let z is the product of x and y.
$\Rightarrow\ \text{z}=\text{xy}\ \Rightarrow\ \text{z}=\text{x}(24-\text{x})\ \ [\text{From eq.(i)}]$
$\Rightarrow\ \text{z}=24\text{x}-\text{x}^2$ $\Rightarrow\ \frac{\text{dz}}{\text{dx}}=24-2\text{x}\text{ and }\frac{\text{d}^2\text{z}}{\text{dx}^2}=-2$
Now to find turning point, $\frac{\text{dz}}{\text{dx}}=0\ \Rightarrow\ 24-2\text{x}=0\ \Rightarrow\ \text{x}=12$
$\text{At }\text{x}=12, \ \frac{\text{d}^2\text{z}}{\text{dx}^2}=-2 \ \ [\text{Nrgative}]$
$\therefore$ x = 12 is a point of local maxima and z is maximum at x = 12.
$\therefore$ From eq. (i), y = 24 = -12 = 12
Therefore, the two required numbers are 12 and 12.

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