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Find the shortest distance, d, from the point (10, 0, -6) to the plane x + y + z = 6. d = Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x, y) = x3 - 108xy + 216 y3 local maximum value(s) local minimum value(s) saddle point(s) (x, y, f) =

Explanation / Answer

d = (a x0+ b y0 + c z0)/ ( a^2 + b^2 + c^2 )^1/2

putiing values,

d = (10 +0 - 6-6)/( 1+1+1)^1/2

d = 2/(3^1/2)

d = 1.1547 unit

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f(x,y ) = x^3 -108 xy + 216 y^3

differentiating it w r t x and putting it zero,

0 = 3 x^2 - 108 y,

x^2 = 36 y ......(1)

now,

differentiating it w r t yand putting it zero

0 = -108 x + 648 y^2,

x = 6 y^2.......(2)

using 1 and 2 ,

36 y^4 = 36 y,

solving it,

y = 0,

y = 1,

so the critical points are,

(0,0) and (6,1)

now checking for , fxx and fyy

A =fxx = 6x ...and at both point it is +ve

now ,

C= fyy = 1296 y

B = fxy = -108

at (0,0),

A,B, C are zero

hence ,

AC- B^2 = 0,

hence,

test is inclusive at this point.

at (6,1),

A = 36 ,

B = -108

C = 1296

AC- B^2 = 36*1296 - 108^2

so ,

AC-B^2 = 34992

hence ,

AC-B^2 is +ve

and A is also positive,

so ,

(6,1) is a point of minimum.

and there is no saddle point.

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