Question

Suppose we are looking for the point on the plane x + 2y + z = 5 closest to the point (2,3,0). Which of the following approaches DOES NOT lead to the answer? = 0 Solve the system of equationsJ 2(x - 2) - 2(5 - x - 2y) | 2(y-3) - 4(5 - x - 2y) = 0 Find the intersection point of the line r(t) = (2+ t, 3 + 2t, t) with the given plane. 2(x 2 2(y 3) = 0 = 0 Solve the system of equations 22 = 0 2( 2) 2(y 3 = > =22 Solve the system of equations 2z x 2yz= 5 Question 5 2 pts Consider the function f(x,y, z) = 5 + (x - 6)2+ (y- 2)2 + (z - 4)2 on the sphere a2 y2 + z2 = 14. Then, , and the absolute minimum is 25 (V14- 6)2 the absolute maximum is 25+(-/14 - 6)2 4-6) ( 14 - 4)2 - 2)2 + the absolute maxi mum is 5 + .and the absolute minimum is 3 (- - 6)? +(- 14 14 14 - 2)2 ( 4)2 5 1 the absolute maximum is 131, and the absolute minimum is 19. the absolute maximum is 14, and the absolute minimum is ) Consider the region D in the ry-plane bounded by the parabola y = 4- x2 and the a-axis. Consider the function f(x, y)( 1)2 2y. _ The absolute maximum value of f(x, y) for (x, y) in D is 8 The absolute minimum value of f(x, y) for (x, y) in D is 0 Question 3 3 pts Consider the function f(x,y) = 2x2 + (y - 1)2 and the disk D : x2 + y2

Answer 1

Question 5 Consider the function f(x, y, z) 5+ ( 6)2 + (y-2)2 +(z-4)2 on the sphere a2 +y+14. Then, the absolute maximum is 25 +(-V14 6)2 and the absolute minimum is 25 +(V14-6)2 the absolute maximum is 5-+(-62+-2)2 + -4and the absolute minimum is 2 +(-- 5+( the absolute maximum is 131. and the absolute minimum is 19 the absolute maximum is 14, and the absolute minimum is 0 Ans. f(x, y,z)5+(x-6 +(y-2+(z-4) g(xy,z) x +z-14 F(xy.z.2)-f(xy.z)+ ig(xy.2)= 5+(x-6) +(y-2+(z-4)+2(x+y- 14 By taking the partial derivatives of F(x.y.z, A)5+ (x-6) +(-2)+(z-4) +2(x+y+z-14), equate to zero and solve for x, y, and z, this gives

6 F(x y,z) 2(x-6)+ 22x =0x : 1+A 2 y2 F.(x, y.z) 2(y-2)+22y 0 4 F (x,y,z) 2(z-4)+21z 0z= 1+A 2 6 36+4+16-14(1+2 = 41+2=t2 2 4 F (x y,z) x +y +z-14 14 + 12) (1+A) 1+A 1A 6 2 4 this gives, Substitute 1+2 = #2 into X =- y 12 Z = 1A 12 6 6 ±3, y 2 2 -t- 2 4 4 =±-=2 2 1 , z 1A 1A 2 12 Substitute x-13,y = t1, z - # 2 into f (x y,z)5+(x-6)-+(y-2)(z-4), this gives, f(xy,z)=5+(x-6 +(y-2+(z-4) f(x, y,z) 5+(3-6) +(1-2)+(2-4) = 5+9+ 1+ 4= 19 Minimum f(x, y,z) 5+(-3-6)+(-1-2) +(-2-4) 5+81+9+ 36=131 Maximun The correct option: the absolute maximum is 131. and the absolute minimum is 19.