Extrema of functions of two variables

Mathematics, winter semester 2018/2019

28.11.2018

Problems1. Find the second partial derivatives of the function (a) f (x, y) = y5− 3xy;

(b) f (x, y) = x−yx+y ; (c) f (x, y) = ln

(x2 + 3y

).

2. Find the local maximum and minimum values and saddle point(s) of thefunction. If you have three-dimensional graphing software, graph the func-tion with a domain and viewpoint that reveal all the important aspectsof the function. (a) f (x, y) = 9 − 2x + 4y − x2 − 4y2; (b) f (x, y) =(1 + xy)(x+ y); (c) f (x, y) = xy + 1

x + 1y ; (d) f (x, y) = ey

(y2 − x2

).

3. Find the absolute maximum and minimum values of on the set D. (a)f (x, y) = 1+4x−5y, D is the closed triangular region with vertices (0, 0) ,(2, 0) and (0, 3) ; (b) f (x, y) = x2+y2+x2y+4, D =

{(x, y) ∈ R2 : |x| ≤ 1 and |y| ≤ 1

};

(c) f (x, y) = xy2, D ={(x, y) ∈ R2 : x ≥ 0, y ≥ 0 and x2 + y2 ≤ 3

}.

4. Find the dimensions of the rectangular box with largest volume if the totalsurface area is given as 64 cm2.

5. (*)The base of an aquarium with given volume V is made of slate and thesides are made of glass. If slate costs five times as much (per unit area) asglass, find the dimensions of the aquarium that minimize the cost of thematerials.

Answers1. (a) f (x, y) = y5 − 3xy; fx,x(x, y) = 0, fx,y(x, y) = fy,x(x, y) = −3,

fy,y(x, y) = 20y3;

(b) f (x, y) = x−yx+y ; fx,x(x, y) = −

4y(x+y)3 , fx,y(x, y) = fy,x(x, y) =

2(x−y)(x+y)3 ,

fy,y(x, y) =4x

(x+y)3 ;

(c) f (x, y) = ln(x2 + 3y

); fx,x(x, y) = −

2(x2−3y)(x2+3y)2

, fx,y(x, y) = fy,x(x, y) =

− 6x(x2+3y)2

, fy,y(x, y) = − 9(x2+3y)2

.

2. (a) f (x, y) = 9− 2x+4y−x2− 4y2; f(−1, 1

2

)is the only local maximum

value of f , no local minimum values neither saddle points;

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(b) f (x, y) = (1+xy)(x+ y); (x, y) = (1,−1) and (x, y) = (−1, 1) are theonly saddle points, no local minimum neither local maximum values;

(c) f (x, y) = xy+ 1x + 1

y ; f(1, 1) is the only local minimum value, no localmaximum values neither saddle points;

(d) f (x, y) = ey(y2 − x2

); f(0,−2) is the only local maximum value,

(x, y) = (0, 0) is the only saddle point, no local minimum values.

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3. (a) f (x, y) = 1 + 4x− 5y, D is the closed triangular region with vertices(0, 0) , (2, 0) and (0, 3) ; the absolute maximum equals f(2, 0) = 9 and theabsolute minimum equals f(0, 3) = −14;(b) f (x, y) = x2 + y2 + x2y + 4, D =

{(x, y) ∈ R2 : |x| ≤ 1 and |y| ≤ 1

};

the absolute maximum equals f(−1, 1) = f(1, 1) = 7 and the absoluteminimum equals f(0, 0) = 4;

(c) f (x, y) = xy2, D ={(x, y) ∈ R2 : x ≥ 0, y ≥ 0 and x2 + y2 ≤ 3

}; the

absolute maximum equals f(1,√2) = f(1,−

√2) = 2 and the absolute

minimum equals f(−1,√2) = f(−1,−

√2) = −2.

4. The greatest volume among all rectangular boxes with the total surfacearea 64 cm2 has the cube with the side equal 4

√2/3 cm.

5. (*) The dimensions of the aquarium that minimize the cost of the materialsare: the base is a square with the side equal 3

√0.4V and the height of the

aquarium is equal 3√6.25V .

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