Let xy= t
then putting the value of x = t/y in the given equation we have,
a2 t4 / y4 + b2 y4 = c6
frm here we have a2 t4 = c6 y4 - b2 y8
for t to be maximum, dt/dy = 0
differentiatin wrt to y, nd then equating it to 0, we have
4 c6 y3 = 8b2 y7
=> y4 = 4c6/ 8b2 = c6/ 2b2
now put this value in the given equation,
a2 x4 = c6 /2
=> x4 = c6 /2a2
thrfore max value of t = fourth root of (c12 / 4a2b2 )
= c3 / fourthroot of (4a2b2)
= c3 / sqrroot of (2ab)
done 
Moderation Team
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