Applications of Derivatives
The sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum when the angle between them is π/3.
I'm also testing Infinite Infinity with this question and seeing what the response time is, so as to know whether it's more practical to use this website or to call up say Rungta and ask him what the solution is.
--Karan
I'm also testing Infinite Infinity with this question and seeing what the response time is, so as to know whether it's more practical to use this website or to call up say Rungta and ask him what the solution is.
--Karan
posted by Karan at 6:56 PM
5 Comments:
The area is given by an Integral of the function describing the triangle. A differential of the integral function which is negative should do the trick...
Not helpful.
let the hypotenuse be y
let the base be x
x + y = C (given)
Area A = 1/2 Base * Altitude
Altitude = (y^2 - x^2)^1/2 (Using the Pyth. Theorem)
so now basically
A^2 = [x^2(y^2 - x^2)] / 4 ....(i)
(we take Area Square to make the calculations easier.. otherwise taking only A will also do)
Now Differentiate (i) and apply Maxima Minima Concepts
you shud end up with A is Max when x = C/3.
now y = C - x
y = C - C/3
y = 2C/3
Base/Hypotenuse (Cosine)= (C/3)/(2C/3) = 1/2
Cos is 1/2 when the angle is 60 or Pi/3 . hence proved
yeah before differentiating change y in terms of x and c (basically y = c-x)
Thanks, that's what I wanted. The full answer.
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