Find the right triangle of largest area given that

Question

the sum of one leg of the triangle and the hypotenuse is 14?

Answer 1

Consider right angled triangle.

Base = b , Opposite = o , Hypotenuse = h

The sum of one leg and the hypotenuse of the triangle is 14

h+b = 14 (Consider that leg is base)

For any right angled triangle ⇒ h² = o²+b² ⇒ o² = h² - b² ⇒ o = √(h² - b²)

 

Area of right angled triangle A = (1/2)bo = (1/2)b√(h² - b²) = (1/2)√(b²h² - b⁴)

A = (1/2)√(b²h² - b⁴)

Apply derivative with respect to b for maximum area.

dA/db = (1/2√(b²h² - b⁴))(2bh²-4b³)

But area is constant ⇒ dA/db = 0

(1/2√(b²h² - b⁴))(2bh²-4b³) = 0

(2bh²-4b³) = 0

h² = 2b²

h = b√2

 

h+b = 14 ⇒ b√2+b = 14 ⇒ (1+√2)b = 14 ⇒ b = 14/(1+√2)

 

h = b√2 ⇒ (14√2)/(1+√2)

o = √[h² - b²]  ⇒ √[(b√2)² - b²] ⇒ √[2b² - b²] ⇒ √[b²] = b

o = b = 14/(1+√2)

Solution : The right angled triangle with maximum area is

Base = 14/(1+√2) , Opposite = 14/(1+√2) , Hypotenuse = (14√2)/(1+√2)