Find the vector and Cartesian equation of the perpendicular drawn from the point (1, -2, 3) to the plane 2x-3y+4z+9=0
Answers
Answer:
answr
search
What would you like to ask?
12th
Maths
Application of Derivatives
Maxima and Minima
Let 2x + 3y + 4z = 9 , x , ...
MATHS
avatar
Asked on December 27, 2019 by
Gaurika Vizhi
Let 2x+3y+4z=9, x, y, z>0, then the maximum value of (1+x)
2
(2+y)
3
(4+z)
4
is
HARD
Share
Study later
ANSWER
Let us consider, f=(1+x)
2
(2+y)
3
(4+z)
4
and g=2x+3y+4z=9
We know that,
Δf=λΔg
2[(1+x)(2+y)
3
(4+z)
4
,3(1+x)
2
(2+y)
2
(4+z)
4
,4(1+x)
2
(2+y)
3
(4+z)
3
]=λ[2,3,4]
⇒2(1+x)(2+y)
3
(4+z)
4
=2λ
3(1+x)
2
(2+y)
2
(4+z)
4
=3λ
⇒4(1+x)
2
(2+y)
3
(4+z)
3
=4λ
λ=(1+x)(2+y)
3
(4+z)
4
=(1+x)
2
(2+y)
2
(4+z)
4
=(1+x)
2
(2+y)
3
(4+z)
3
Here, x,y,z>0
On solving we get, (2+y)(4+z)=(1+x)(4+z)=(1+x)(2+y)
⇒(2+y)(4+z)=(1+x)(4+z)and(1+x)(4+z)=(1+x)(2+y)
⇒2+y=1+x,4+z=2+y
⇒x=1+y,z=y−2
by substituting the x and z value in g we get 2(1+y)+3y+4(y−2)=9
Finally , (x,y,z)=(
3
8
,
3
5
,
3
−1
)
f(
3
8
,
3
5
,
3
−1
)
And the maximum value is (
3
11
)
9