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if x+y=4 and x>=0,y>=0 find the maximum value of x^3 y

if x+y=4 and x>=0,y>=0 find the maximum value of x^3 y

Grade:11

2 Answers

Rinkoo Gupta
askIITians Faculty 80 Points
7 years ago
s=x3y=x3(4-x)=4x3-x4
put ds/dx=0
12x2-4x3=0
4x2(3-x)=0
x=0,3
d2s/dx2=24x-12x2
the value of d2s/dx2at x=3 is 24(3)-12(9)=72-108=-36 <0
so x3y is max at x=3
so max. value of x3y is (3)3(4-3)=27(1)=27
Ans. is 27.
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
so maximum value of x3y is
Thanks & Regards
Rinkoo Gupta
askIITians Faculty
Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans: 27
Sol:
f(x, y) = x^{3}y
x+y=4
f(x) = x^{3}(4-x)
First, find the critical points of ‘f’.
f^{'}(x) = 0
f^{'}(x) = 12x^{2}-4x^{3} = 0
x = 0, 3
f^{''}(x) = 24x-12x^{2}
f^{''}(0) = 24.0-12(0)^{2} = 0
f^{''}(3) = 24.3-12(3)^{2} = 72 - 108 = -36
So, x = 3 is the maxima.
f(3) = 3^{3}(4-3) = 27
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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