A calculus problem by A Former Brilliant Member

By pythagorean theorem, the hypotenuse of the triangle is $\sqrt{80^2+60^2}=\sqrt{10000}=100$

By similar triangles, we have, $\dfrac{y}{a}=\dfrac{60}{80}$ $\implies$ $a=\dfrac{4}{3}y$

By similar triangles again, $\dfrac{y}{100-x-a}=\dfrac{80}{60}$ $\implies$ $y=\dfrac{4}{3}(100-x-a)$

However, $a=\dfrac{4}{3}y$ , so

$y=\dfrac{4}{3}(100-x-\dfrac{4}{3}y)$ $\implies$ $\dfrac{3}{4}y=100-x-\dfrac{4}{3}y$ $\implies$ $\dfrac{25}{12}y=100-x$ $\implies$ $y=\dfrac{12}{25}(100-x)$

$A=xy=x\left(\dfrac{12}{25}(100-x)\right)=\dfrac{12}{25}(100x-x^2)$

Differentiate both sides with respect to $x$ .

$\dfrac{dA}{dx}=\dfrac{12}{25}(100-2x)$

Let $\dfrac{dA}{dx}=0$

$\dfrac{12}{25}(100-2x)=0$ $\implies$ $100=2x$ $\implies$ $\boxed{x=50~feet}$

Let the hypotenuse side of top small the right angle triangle is x. Then the bulding has sides of 4/5 x and 5/3 (60 – x). So the area = 80 x – 4/3 x^2. dA/dx = 80 – 8/3 x = 0, x= 30. So the longest side of the building = 50,