Find the absolute maximum and absolute minimum of the function f(x, y)=2x2−4x+y2−4y+1 on the closed triangular plate bounded by the lines x=0,y=2,y=2xin the first quadrant.

This question is based on the absolute maximum and absolute minimum.

We get this by differentiating the terms.

Given:

f(x,y) =   , bounded by the lines x=0,y=2,y=2x in the first quadrant,bounded by the lines x=0,y=2,y=2x in the first quadrant.

We need to determined the absolute maximum and absolute minimum of the function.

Now, partial differentiating wrt x and y.

Now, point (1, 2) lies on the line y = 2x ; at this point, we get a value of

f(1, 2) = -5 (MIN).

Next, check the points where the boundary lines intersect, which occurs at the points (0, 0), (0, 2), and (1, 2).

Now, find f(0, 0) = 1 (MAX) and f(0, 2) = -3.

Now check on the boundary lines themselves.

If x = 0, then we get,

which has a maximum value of -3 when y = 2 (so we get the same critical point as before).

If y = 2, then we get,

f(x,2) = with a maximum of -5 when x = 1.

If y = 2x, then we get,

f(x,2x) = with the same maximum of -5 when     x = 1.

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First check for the critical points of f by checking where the first-order derivatives vanish.

Notice how the point (1, 2) lies on the line y = 2x ; at this point, we get a value of f(1, 2) = -5 (MIN).

Next, check the points where the boundary lines intersect, which occurs at the points (0, 0), (0, 2), and (1, 2). We already checked the last one. We find f(0, 0) = 1 (MAX) and f(0, 2) = -3.

Now check on the boundary lines themselves. If x = 0, then

which has a maximum value of -3 when y = 2 (so we get the same critical point as before).

If y = 2, then

with a maximum of -5 when x = 1.

If y = 2x, then

with the same maximum of -5 when x = 1.

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

f)

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

a)

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

b)

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

c)

the function given is   y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

d)

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 =   -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e)

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

f)

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

g)

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24

circle area = pi * radius^2

if diameter = 10 then radius = 5

circle area = pi * 5^2

circle area = pi * 25

circle area = 78.5398163397 square units

step-by-step explanation: