SOLUTIONS TO HOMEWORK ASSIGNMENT #5, Math 253

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1 SOLUTIONS TO HOMEWORK ASSIGNMENT #5, Math 53. For what values of the constant k does the function f(x, y) =kx 3 + x +y 4x 4y have (a) no critical points; (b) exactly one critical point; (c) exactly two critical points? Hint: Consider k =0andk 0 separately. Set f x =0an y = 0 to find critical points: f x =3kx +x 4=0 () f y =4y 4=0 () () gives y =. For (), consider k =0andk 0 separately. For k = 0, () becomes x 4 = 0, or x =. So one critical point at (, ). For k 0, use quadratic formula to solve for x. x = ± 4+48k 6k = ± +k 3k So critical points are ( ± +k, ) if they exist. 3k Conclusion: k< /: no critical points. k = /: one critical point (4, ). k> / and k 0: two critical points ( ± +k, ). 3k k = 0: one critical point (, ).. Find and classify all critical points of the following functions. (a) f(x, y) =x 3 y 3 xy +6 f x =3x y =0 () f y = 3y x =0 () () gives y = 3 x. Substituting into () becomes 3 ( 3 x) x = 0, or simplified x(7x 3 +8)=0. Hencex =0or /3. If x =0,thenby()y =0 (0, 0) If x = /3, then by () again y =/3 ( /3, /3). Hence, critical points at (0, 0) and ( /3, /3).

2 Step : apply second derivative test f xx =6x f yy = 6y f xy = At (0, 0), f xx =0,f yy =0,f xy =. So D = f xx f yy (f xy ) = 4 < 0 saddle At ( /3, /3), f xx = 4 < 0, f yy = 4, f xy =. So D => 0 local max Hence, local max at ( /3, /3), saddle point at (0, 0) (b) f(x, y) =x 3 + y 3 +3x 3y 8 f x =3x +6x =0 () f y =3y 6y =0 () We can solve the two equations separately. () gives x = 0 and. () gives y = 0 and. Hence, there are four critical points at (0, 0), (0, ), (, 0), and (, ). Step : apply second derivative test f xx =6x +6 f yy =6y 6 f xy =0 At (0, 0), f xx =6,f yy = 6, f xy =0,soD = 36 < 0 saddle At (0, ), f xx =6> 0, f yy =6,f xy =0,soD =36> 0 local min At (, 0), f xx = 6 < 0, f yy = 6, f xy =0,soD =36> 0 local max At (, ), f xx = 6, f yy =6,f xy =0,soD = 36 < 0 saddle Hence, local max at (, 0), local min at (0, ), saddle at (0, 0) and (, ) (c) f(x, y) = x +y x f x = =0 () (x + y ) y f y = =0 () (x + y ) () gives x = 0 and () gives y = 0. The critical point is at (0, 0). Step : apply second derivative test f xx = (x + y ) x[(x + y )(x)] (x + y ) 4 f yy = (x + y ) y[(x + y )(y)] (x + y ) 4 f xy = x()(y) (x + y ) 3 At (0, 0) f xx = < 0, f yy =, f xy =0,SoD =4> 0 local max Hence, local max at (0, 0)

3 (d) f(x, y) =y sin x f x = y cos x =0 () f y =sinx =0 () () gives x = nπ for all n Z, i.e. integers. Substituting to () gives ±y =0,or y = 0. The critical points are (nπ, 0) for all n Z. Step : apply second derivative test f xx = y sin x f yy =0 f xy =cosx At all (nπ, 0), f xx =0,f yy =0,f xy = ±, so D = < 0 saddle Hence, saddle points at (nπ, 0) for all n Z 3. Suppose f(x, y) satisfies the Laplace s equation f xx (x, y)+f yy (x, y) = 0 for all x and y in R. If f xx (x, y) 0 for all x and y, explain why f(x, y) must not have any local minimum or maximum. Since the second derivatives exists, the first derivatives must be continuous and f(x, y) must be differentiable. Also, since there is no boundary on R, local max/min must occur at critical points. Suppose there is a critical point, then by second derivative test, D = f xx f yy fxy.but f xx + f yy =0 f yy = f xx. It follows that D = fxx fxy < 0whenitisgiventhat f xx 0. Therefore all critical points are saddle points. 4. Find all absolute maxima and minima of the following functions on the given domains. (a) f(x, y) =x 4x + y 4y + on the closed triangular plate with vertices (0, 0), (, 0), and (, ) Step : find interior critical points f x =4x 4=0 () f y =y 4=0 () () gives x =. () gives y =. Critical point at (, ), but not in region. Step : find boundary critical points and endpoints Bottom side y =0 f(x, 0)=x 4x +. dy =4x 4=0 x =. Critical point at (, 0) Right side x = f(,y)=8 8+y 4y +=y 4y +. =y 4=0 y =. Critical point at (, ). dx Hypotenuse y = x f(x, x) =x 4x + x 4x +=3x 8x + =6x 8=0 x =4/3. So y =4/3. Critical point at (4/3, 4/3). dx Together with the endpoints of all sides (0, 0), (, 0), (, ). 3

4 Step 3: compare the values of f(x, y) f(, 0) = f(, ) = 3 f(4/3, 4/3) = 3/3 absolute min f(0, 0) = absolute max f(, 0) = absolute max Hence, abs max at f(, 0) = f(0, 0)=,absminatf(4/3, 4/3) = 3/3 (b) f(x, y) =x + xy +3x +y + on the domain D = {(x, y) x y 4} Step : find interior critical points f x =x + y +3=0 () f y = x +=0 () () gives x =. Substituting to () gives y =. Critical point at (, ) but not in region. Step : find boundary critical points Top side: y =4 f(x, 4) = x +4x +3x +8+=x +7x +0 dx =x +7=0 x = 7/ but not in region Parabola: y = x f(x, x )=x + x 3 +3x +x +=x 3 +3x +3x + dx =3x +6x +3=3(x +) =0 x =, then y =( ) =. Critical point (, ). Together with the endpoints of the two sides (, 4), (, 4). Step 3: Compare the values of f(x, y) f(, ) = f(, 4) = 0 absolute min f(, 4) = 8 absolute max Hence, absolute min at f(, 4) = 0, absolute max at f(, 4) = 8 (c) f(x, y) =x +3y 4x 5 on the domain D = {(x, y) x + y 6}. Step : find interior critical points f x =4x 4=0 () f y =6y =0 () () gives x =. () gives y = 0. Critical point (, 0). Step : find boundary critical points Rewrite the boundary y =6 x or y = ± 6 x, which the endpoints are (4, 0) and ( 4, 0). Then f becomes f =x +3(6 x ) 4x 5= x 4x dx = x 4=0 x =, y =6 ( ) y = ± Critical points at (, ) and (, ). Step 3: compare the values of f(x, y) f(, 0) = 7 absolute min f(4, 0) = 4

5 y y D y 0 D 0.5 D x x x Figure : Q4(a) Figure : Q4(b) Figure 3: Q4(c) f( 4, 0) = 43 f(, ) = 47 absolute max f(, ) = 47 absolute max Hence, abs min at f(, 0) = 7, abs max at f(, ) = f(, ) = Use Lagrange multipliers to find the maximum and minimum values of the following functions subject to the given constraint(s). (a) f(x, y) =xy subject to x +y = Step : Find critical points on constraint f(x, y) =xy, f x = y, f y = x g(x, y) =x +y =,g x =x, g y =4y y =λx () x =4λy () x +y = (3) Substituting () into () gives x =4λ(λx), or x(8λ ) = 0 x =0or λ = ± 8. For x =0,() gives y = 0, but contradicts with (3). No solution in this case. For λ =/ 8, () gives x = y. Substituting into (3) gives y +y = y = ±/. So x = ±/. Critical points at (/, /), ( /, /). For λ = / 8, () gives x = y. Substituting into (3) gives y +y = y = ±/. So x = /. Critical points at ( /, /), (/, /). Step : Compare the values of f(x, y) f(/, /) = / absolute max f( /, /) = / absolute max f( /, /) = / absolute min f(/, /) = / absolute min Hence, abs max at f(/, /) = f( /, /)=/, abs min at f( /, /) = f(/, /) = /. (b) f(x, y, z) =xy + z subject to y x =0andx + y + z =4 5

6 Step : Find critical points on constraints f(x, y) =xy + z, f x = y, f y = x, f z =z g(x, y) =y x =0,g x =, g y =,g z =0 h(x, y) =x + y + z =4,h x =x, h y =y, h z =z y = λ +µx () x = λ +µy () z =µz (3) y x =0 (4) x + y + z =4 (5) (4) gives y = x. Substitute into () and () x = λ +µx x = λ +µx (a) (a) (a) (a) givesλ = 0. () and () becomes x =µx y =µy (b) (b) (b) and(b) gives either x = y =0orµ =/. For x = y =0,(5) gives z = ±, and (3) gives µ =. Critical points at (0, 0, ) and (0, 0, ) For µ =/, (3) gives z = 0. (5) becomes x + x =4 x = ±, then y = ±. Critical points at (,, 0) and (,, 0) Step : Compare the values of f(x, y) f(0, 0, ) = 4 absolute max f(0, 0, ) = 4 absolute max f(,, 0) = absolute min f(,, 0) = absolute min Hence, absolute max at f(0, 0, ) = f(0, 0, ) = 4, absolute min at f(,, 0) = f(,, 0) =. 6