Question

Find an equation of the plane that passes through the point (1, 3, 4) and cuts off the smallest volume in the first octant.

Answer 1

Answer:

12x +4y + 3z=36

Step-by-step explanation:

The equation of plane is given by

z-zo = a(x-xo) + b(y-yo)

pass through (1,3,4)

Z -4 = a(x -1) +b(y-3)

The question is asking us to optimize a and b. To minimize the volume V both a and b should be negative as the normal vector should be towards the negative x and y direction so that a finite tetrahedron can be formed in the first octant.

we need x , y and z intercepts o define volume

x intercept( y, z =0) = $\frac{a+3b-4}{a}$

y intercept (x, z =0) = $\frac{a+3b-4}{b}$

z intercept ( x, y =0) = -(a+3b-4)

Base = $\frac{(a+3b-4)^2}{2ab}$

Volume = $\frac{1}{3}*base*height$

Volume(a, b) = $\frac{-(a+3b-4)^3}{6ab}$

now we differentiate partially in terms to a and b the volume to minimize and get a and b.

ΔV(a, b) = $\frac{-1}{6}(\frac{3(a+3b-4)^2ab-b(a+3b-4)^3}{a^2b^2}$ ,$\frac{-1}{6}(\frac{9(a+3b-4)^2ab-a(a+3b-4)^3}{a^2b^2}$  = 0

Taking the first part of differential it will give

b(a+3b-4) [3a -(a+3b -4)] =0

(a+3b-4) $\neq 0$ because the volume will become zero if this becomes true

2a -3b = -4  ..................(1)

similarly the second part of the differential will give

a-6b=4 ................(2)

on solving 1 and 2 we get

a = -4 and b = -4/3

so the equation will be

Z -4 = -4(x -1) - 4/3*(y-3)

final equation

12x +4y + 3z=36