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Find the integer value of x for which the area of a triangle with sides 9, 40x and 41x is maximized?

+1 vote
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Find the integer value of x for which the area of a triangle with sides 9, 40x and 41x is maximized?
posted Nov 1 by anonymous

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1 Answer

+1 vote

Area of triangle = sqrt(p×(p-a)×(p-b)×(p-c)) where p = perimeter/2
p = (9 + 40x + 41x)/2 = (9 + 81x)/2
Area^2 = ((9 + 81x)/2)×((81x - 9)/2)×((9+x)/2)×((9-x)/2)
Area = 9/4×sqrt(6562x^2 - 81x^4 - 81)
d(Area)/dx = -(729x^3 - 29529x)/2×(sqrt(-81x^4 + 6562x^2 - 81)) = 0
To get maxima (because minima is area = 0 for x=0 in this case)
x = +/- 6.3644
Therefore for x = 6 we have maximum triangle area with sides
9, 240, 246.

answer Nov 1 by Tejas Naik
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