1Question 1
Assuming x,y,z≥0
for maxima of given expression, it is clearly visible that if z is maximum the product will be maximum, also y>x.
Hence:
z>y>x
given x+y+z=6;
x=1;y=2;z=3
thus, the expression is equal to: 1*4*27=108
1) find maxima of xy2z3 provided : x + y + z = 6.
2) find maxima of (x-3)2 + (y-2)2 provided : (x+3)2 + (y+2)2 = 4
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9 Answers
1
21U can easily use lagranges multiplier for both the Qs.....it jus kills such types of Q's
39Wiki mein bohut ajeeb sa diya hua hai...eragon can you kindly explain? We would be indebted to you if you took out some of your precious time to help us illiterate people. We're scared of such foreign terms.
3411) Well known: Given that
x+y+z = x + \frac{y}{2} + \frac{y}{2} + \frac{z}{3} + \frac{z}{3} + \frac{z}{3} = 6
now, use AM-GM to obtain the max as 22X33 = 108
2) You are being asked what is the maximum distance of a point on the circle (x+3)^2 + (y+2)^2 = 4 from the point (3,2). From triangle inequality the max of the expression is 4(\sqrt {13} + 1)^2
21ok....
Let f(x,y)=(x-3)2+(y-2)2
g(x,y)=(x+3)2+(y+2)2=4
so u hav to maximize f(x,y) subject to the constraint (x+3)2+(y+2)2=4
now we intoduce a lagrangian function
\wedge (x,y,\lambda )=f(x,y)+\lambda (g(x,y)-4)
\wedge (x,y,\lambda )=f(x,y)+\lambda (g(x,y)-4)\\ \Rightarrow \wedge (x,y,\lambda )=(x-3)^2+(y-2)^2+\lambda \left ( (x+3)^2 + (y+2)^2 - 4 \right )\\ \textup{Now partially differentiate} \wedge (x,y,\lambda ) \textup{wrt x ,y and } \lambda \textup{and eqaute it to 0}\\ \frac{\delta \wedge}{\delta x}=2(x-3)+\lambda. 2.(x+3)=0...........i\\ \frac{\delta \wedge}{\delta y}=2(y-2)+\lambda .2.(y+2)=0...........ii\\ \frac{\delta \wedge}{\delta \lambda }=(x+3)^2 + (y+2)^2 - 4=0........iii\\
now find x in terms of λ from eq i
similarly y in terms of λ from eq ii
and eqaute x and y in terms of λ in eq iii
so that u get values of λ(may be more than 1)
now find f(x,y) in terms of λ and substitute the value of λ u found in f(λ) to find max and min value of f(λ) and susequently f(x,y)
Note-this is a method which u can jus blindly follow without thinking much
there may be short ways also as prophet sir did.....but not always der of shorter method and in those u can use this
here in this q u get
\lambda =-\sqrt{13}-1 \; and\; \sqrt{13}-1 \\ \textup{ here for} \lambda =-\sqrt{13}-1 \\ \textup{we get the max value of f(x,y)} \textup{and we hav } f(\lambda )=\frac{52\lambda ^2}{(1+\lambda )^2}\\ \textup{jus substitute } \lambda =-\sqrt{13}-1 \textup{for max value of} f(\lambda )
This may look lenghty but its actually not that lenghty.
But use this only when nothing else is coming into ur head ;)
