Find the volume of the largest rectangular box in the first octant with three fa

Question

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 3.

Explanation / Answer

Soln: x+2y+3z=3 => z= (3-x-2y)/3 let V=f(x,y)= XYZ V= f(x,y) = xy( (3-x-2y)/3 ) if the volume is maximum, Fx ,Fy =0. (Derivative w.r.t x & y is ZERO) A) fx= 1/3 [ Y.(3-2x-2y) ]=0 => y=0 or y=1.5- x B) fy =1/3 [x.(3-x-4y)] =0 substitute y=0 & y=1.5 -x into B, at y=0 we find ,x=0 & x=3 and at y=1.5-x we find , x=0 & x=1 AND so AT X=0 => y=1.5 AND at x=1 => Y=0.5 USE CRITICAL POINTS: (0,0) (3,0) ,(0,1.5) ,(1,0.5) D=D(a,b) = Fxx(a,b) * Fyy(a,b) - [ Fxy (a,b ] ^2 Fxx=-2y/3 Fyy=-4x/3 Fxy= 1/3[ 3-2x-4y ] at (0,0) =====> D= -1 at (3,0) =====> D= -1 at(0,.15) ======> D= -1 at (1, 0.5) ======> Fxx =-2/3 Fyy =-4/3 Fxy =-1/3 D=7/9 We can see that F(1,0.5) produces maximum values without computing second derivative thus z=(3-x-2y)/3 =1/3 V= (1)(0.5)(1/3) =1/6 is the volume. ANS =1/6 Please RATE:) Thanking You

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