IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

pottermania1990

find the max. value of xy when

2 4 2 4 6

a x + b y = c x,y>0

magiclko

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

pottermania1990

srry thats not d ans.

magiclko

then wat is the answer... ???

pottermania1990

3

c

_________

sqrrt(2ab)

c cube divided by square root of 2ab

magiclko

i got the mistake, i took c square instead of c to power 6...

iitkgp_bipin

Let xy = z

a²x⁴ + b²y⁴ = c⁶
a²x⁴ + b²(z/x)⁴ = c⁶
z⁴ = (a²x⁸ - c⁶x⁴)/b²

Now differentiate both sides wrt x and put dz/dx = 0 for maxima.
You will get x⁴ = c⁶/2a²
Putting this in the above equation : y⁴ = c⁶/2b²

Hence x⁴y⁴ = (c⁶/2a²)(c⁶/2b²)

xy = c³/(2ab)^1/2

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