Find two numbers whose mean proportion is 36 and the third proportional is 288.

Let the two numbers be x and y.

Thus, 36 is the mean proportion between x and y.

$\Rightarrow \dfrac{x}{36} = \dfrac{36}{y} \\[1em] \Rightarrow xy = 36^2 \\[1em] \Rightarrow xy = 1296 \\[1em] \Rightarrow x = \dfrac{1296}{y}\text{…..(1)}$

The third proportional for x and y is 288.

$\Rightarrow \dfrac{x}{y} = \dfrac{y}{288} \\[1em] \Rightarrow y^2 = 288x \text{ ……….(2)}$

Substituting value of x from equation (1) in (2), we get :

$\Rightarrow y^2 = 288 \times \dfrac{1296}{y} \\[1em] \Rightarrow y^3 = 373248 \\[1em] \Rightarrow y = \sqrt[3]{373248} = 72.$

Substituting the value of y in equation (1), we get :

$\Rightarrow x = \dfrac{1296}{y} \\[1em] \Rightarrow x = \dfrac{1296}{72} \\[1em] \Rightarrow x = 18$

Hence, the two numbers are 18 and 72.