Find the minimum distance from the point to the plane x - y + z = 12. (0. 0. 0)

Question

Find the minimum distance from the point to the plane x - y + z = 12. (0. 0. 0) A home improvement contractor is painting the walls and ceiling of a rectangular room The volume of the room is 875.00 cubic feet. The cost of wall paint is $0.08 per square foot and the cost of ceiling paint is $0.14 per square foot. Find the room dimensions that result in a minimum cost for the paint. Sides ft. height ft What is the minimum cost for the paint? () $ A trough with trapezoidal cross sections is formed by turning up the edges of a 30-inch-wide sheet of aluminum (see Figure) Find the cross section of maximum area.

Explanation / Answer

1)given plane x-y+z=12

point on plane is (x,y,z)

distance from origin, d =[(x-0)2+(y-0)2+(z-0)2]

d =[x2+y2+z2]

distance is minimim if x2+y2+z2 is minimum

let f(x,y,z)=x2+y2+z2

we have x-y+z=12 =>y =x+z-12

f(x,z)=x2+(x+z-12)2+z2

f(x,z)=2x2+2xz-24x+2z2-24z+144

for critical points fx =0 , fz =0

fx=4x+2z-24, fz=2x+4z-24

solving 4x+2z-24=0, 2x+4z-24=0

4x+2z=24, 2x+4z=24

8x+4z=48,2x+4z=24

8x+4z-2x-4z=48-24

6x=24

x=4

2x+4z=24

2*4 +4z=24

4z =16

z =4

x-y +z=12

4-y+4=12

y =-4

critical point is (4,-4,4)

fx=4x+2z-24, fz=2x+4z-24

fxx=4 ,fzz=4,fxz=2

D=fxxfzz-fxz2=4*4 -22 =12

D>0, fxx>0

so distance is minimum when point on plane is (4,-4,4)

minimum distance =[42+(-4)2+42]

minimum distance =43

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