Question

an t. Determine Equation for the plane that passes through the point cl, 3, 9 ) and ents off the smallest volume ju the first octant.

Answer 1

The equation of a plane is given by 2-20= a (x-4)=6 Cy=%). since the point (1,3,9) 8 given in the question, we will Substitute them for (no, yo, 2o) 2-9 = a (x-1) + b (4-3) a and b The question ļ essentially asking us to optimize in order to optimize the volume & as a functim of a and b 3 bta-9 x- intercept (set y=2=0); y- intheept (set, n=2=0): 3 bt a-9 b z-intercept (set, n=y=0) : -a- 3b+q (3b7a-g). La basex Vs height x 12x X(-a-36+9) 3 ab 3 ( 3h+a-gy} Ve Now let's set gradien of X, 8V = 0 and solve fr a&b 6 ab 3(354a-9) ab 1(3642-9) 6 (3bta-gay- a(36+4-98 Dv (a,b) = 6a2b2 6a²b2 co

b(3679-9) CB 640-9)- 3a] => 9 (36+ar -9) [(367a-9) –Gb] 2 D where arbto Note that we cannot solue for 3bta-q=0 because that will lead to y=0, which is impossible as the plane cuts The positive volume from the first quadrant! 3bta-9-3 a 5o = -2a + 3b-920 3b + a-9-66 =o) a-3b- 950 -a 18 んこ --18 18 ta ab = 21 lbig (3h+a-9) Now, X(-18,-1) = (27–18–03 6 ab 6X-9x HB V = 162 Cubic units eqn of plane. z - - - 13 (x-1)= 1 (-3) 3 18n tay +2=54 Ang.