# What is the largest possible area for a right triangle with hypotenuse of 15cm, and what are its dimensions? (using calculus)?

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• 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

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• 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

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• 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

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• The largest area is when the legs are each 15 / sqrt(2).

• david
Lv 7
4 weeks agoReport

true ... but how do you use calculus to prove it?

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