help@askiitians.com

+91-87964 74404

Grade 11Discuss with Askiitians Tutors

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

moidin afsan

12 Years agoGrade 11

2 Answers

Rinkoo Gupta

Approved Tutor Answer12 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

Approved Tutor Answer12 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

Please send the full solution as soon as possible 🙏

discuss with askiitians tutors

1 Answer Available

Last Activity: 3 Years ago

I got 80% in jee mains and am from sc category . Will i get a good NIT

discuss with askiitians tutors

1 Answer Available

Last Activity: 4 Years ago

Looking for an integrated JEE+NEET Coaching Classess for 2 years along with an ongoing normal syllabus related classes followed by regular revision tests, please.
Thanks
Gopinath [F/O Class 10^th Student-Qatar]

discuss with askiitians tutors

1 Answer Available

Last Activity: 4 Years ago

Ntse coaching for class 9 ncert student, fee and schedule

discuss with askiitians tutors

1 Answer Available

Last Activity: 4 Years ago

Can a class 9 student take admission in the course, 2- what is the duration and mode of full course ie how many student in one class.

discuss with askiitians tutors

1 Answer Available

Last Activity: 4 Years ago

View all Questions