Calculus: Smallest hypotenuse in a right-angled triangle with a known area?
If you have a right-angled triangle with area = S, which one would have the smallest hypotenuse?
Let x and y be the legs of the right triangle. Then the hypotenuse z satisfies
z² = x² + y².
Since z>0, minimizing z² will minimize z.
The area of the triangle is
S = (1/2)xy
→
2S/x = y
→
z² = x² + (2S/x)².
So find a minimum for the function
f(x) = x² + (2S/x)²:
f’(x) = 2x + 4S^2 (-2 x^(-3))
= 2x – 8S² x^(-3).
This does not exist when x=0, but if x=0 we don’t have a triangle. Set to 0:
0 = 2x – 8S² x^(-3)
Multiply through by x^3:
0 = 2x^4 – 8S²
0 = x^4 – 4S²
4S² = x^4
[4S²]^(1/4) = x
√(2S) = x.
(You need to perform the second derivative test to show x = √(2S) gives a minimum.)
Then y = 2S/x = 2S/√(2S) = √(2S),
and
z² = (√(2S))² + (√(2S))²
= 2S + 2S
= 4S
–>
z = 2√S is the smallest hypotenuse.
Hope this helps.
♣ ♣
Let x and y be the legs of the right triangle. Then the hypotenuse z satisfies
z² = x² + y².
Since z>0, minimizing z² will minimize z.
The area of the triangle is
S = (1/2)xy
→
2S/x = y
→
z² = x² + (2S/x)².
So find a minimum for the function
f(x) = x² + (2S/x)²:
f’(x) = 2x + 4S^2 (-2 x^(-3))
= 2x – 8S² x^(-3).
This does not exist when x=0, but if x=0 we don’t have a triangle. Set to 0:
0 = 2x – 8S² x^(-3)
Multiply through by x^3:
0 = 2x^4 – 8S²
0 = x^4 – 4S²
4S² = x^4
[4S²]^(1/4) = x
√(2S) = x.
(You need to perform the second derivative test to show x = √(2S) gives a minimum.)
Then y = 2S/x = 2S/√(2S) = √(2S),
and
z² = (√(2S))² + (√(2S))²
= 2S + 2S
= 4S
–>
z = 2√S is the smallest hypotenuse.
Hope this helps.
♣ ♣
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