Distance between projection points on the legs of right triangle (solution by Calculus)

By ratio and proportion
$\dfrac{y}{x} = \dfrac{30}{40}$

$y = \frac{3}{4}x$
 

By Pythagorean theorem
$d = \sqrt{(40 - x)^2 + y^2}$

$d = \sqrt{(40 - x)^2 + (\frac{3}{4}x)^2}$

$d = \sqrt{(40 - x)^2 + \frac{9}{16}x^2}$
 

For minimum length of d, differentiate then equate to zero
$\dfrac{dd}{dx} = \dfrac{2(40 - x)(-1) + \frac{9}{8}x}{2\sqrt{(40 - x)^2 + \frac{9}{16}x^2}} = 0$

$-2(40 - x) + \frac{9}{8}x = 0$

$\frac{25}{8}x = 80$

$x = 25.6 \, \text{ cm}$
 

Distance of D from side AB for minimum length of d
$y = \frac{3}{4}(25.6)$

$y = 19.2 \, \text{ cm}$           answer
 

The same problem was solved by Geometry alone. See the solution here: Distance between projection points