What is the largest possible area for a right triangle with hypotenuse of 15cm, and what are its dimensions? (using calculus)?
- billrussell42Lv 74 weeks ago
let a and b be the other two sides
area = (1/2)ab
a² + b² = 15²
a = √(225 – b²)
area = (1/2)b√(225 – b²)
take derivative, set equal to 0 and solve for b
- davidLv 74 weeks ago
legs are x and y
x^2 + y^2 = 15^2
y = sqrt[225 - x^2]
A = (1/2)xy
A(x) = (x/2)*sqrt[225 - x^2]
A' = (x/2)(1/2)(-2x)/sqrt[225 - x^2] + (1/2)sqrt[225 - x^2] = 0
(-x^2)/2 + (1/2)(225 - x^2) = 0
x^2 + x^2 = 225
x^2 = 225/2
x = 15/sqrt2 <<< do a little algebra and y = same = 15/sqrt2
- PuzzlingLv 74 weeks ago
Intuitively the answer is when you have a 45-45-90 triangle where the legs are equal and have a value of 15/√2.
But if you want to be a little more rigorous you can resort to calculus.
Let a be the measure of one leg.
Let b be the measure of the other leg.
By the Pythagorean Theorem:
a² + b² = 15²
a² + b² = 225
a² = 225 - b²
a = √(225 - b²)
The formula for the area of a right triangle is half the product of the legs.
A = ½ab
Substitute in for a and you'll have a function for the area, based on the value of b.
f(b) = ½√(225 - b²)b
Take the derivative:
f'(b) = ½(225 - 2b²)/√(225 - b²)
Set that to zero. We only really care about the numerator:
225 - 2b² = 0
2b² = 225
b² = 15² / 2
b = √(15²/2)
b = 15/√2
And you can then find out that a = 15/√2
Both legs are 15/√2
- MorningfoxLv 74 weeks ago
The largest area is when the legs are each 15 / sqrt(2).