Find the value(s) of the parameter 'a' (a>0) for each of which the area of the figure bounded by the straight line, y =a2-ax1+a4 and the parabola y = x2+2ax+3a21+a4 is the greatest.
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3 Answers
62first find the points where the two curves are equal...
so a2-ax=x2+2ax+3a2
so 0=x2+3ax+2a2
Hence, x=-a and x=-2a are teh roots
now we need to find the integral.. between -2a and -a
f(x)-g(x) = x2+3ax+2a21+a4
INtegral of this will be x3/3+3/2ax2+2a2x1+a4
applying limits from -2a to -a we get
(-8a3/3+3x2a.a2+2a2(-2a))-(-a3/3+3/2aa2+2a2a)1+a4
(1/6a3)1+a4
=(1/6)a+a-3
maximizing the above is same as minimizing a+a-3
derivative is 1-3a-4
double derivative is +ve...
a=31/4
Make sure i have not made any calculation mistake.. bcos there were a few of those involved...
