The company produces two types of goods in quantities of x and y, with market prices of €40 and 80€, respectively. If the production cost is given by function C(x, y) =2x^2+5y^2+120 and is not exceeding €250. What is the max profit obtained?

€ 270

Step-by-step explanation:

Since the production cost C(x,y) = 2x² + 5y² + 120 is less than or equal to 250, we have  2x² + 5y² + 120 ≤ 250

The selling price S(x,y) = 40x + 80y

The profit P(x,y) = S(x,y) - C(x,y) = 40x + 80y - 2x² - 5y² - 120

Using the principle of lagrange multipliers, we want to maximize the profit P(x,y) under the condition that C(x.y) ≤ 250.

So, dP/dx = 40 - 4x , dC/dx = 4x, dP/dy = 80 - 10y , dC/dy = 10y

dP/dx + λdC/dx = 0

40 - 4x + 4λx = 0  (1)

4λx = 4x - 40

λ = (x - 10)/x

dP/dy + λdC/dy = 0

80 - 10y + 10λy = 0 (2)

substituting λ into (2), we have

80 - 10y + 10(x - 10)y/x = 0

multiplying through by x, we have

80x - 10xy + 10xy - 100y = 0

80x - 100y = 0

80x = 100y

x = 100y/80

x = 5y/4

substituting x into C(x,y) ≤ 250, we have

2(5y/4)² + 5y² + 120 ≤ 250

25y²/8 + 5y² + 120 ≤ 250

25y² + 40y² + 960 ≤ 2000

65y² ≤ 2000 - 960

65y² ≤ 1040

y² ≤ 1040/65

y² ≤ 16

y ≤ ±√16

y ≤ ± 4 since its quantity, we take the positive value.

So x = 5y/4 = 5(± 4)/4 = ± 5

So, x ≤ ± 5

For the maximum value for the profit, P(x,y), we take the maximum values of x and y which are x = 5 and y = 4. Substituting these values into P(x,y), we have

P(5,4) = 40(5) + 80(4) - 2(5)² - 5(4)² - 120

= 200 + 320 - 50 - 80 - 120

= 520 - 250

= 270

So, the maximum profit obtained is € 270

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab  in the y coordinate 

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

  x coordinate 

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

  y coordinate 

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

  x  coordinate 

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

  y  coordinate 

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

  x  coordinate 

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

  y  coordinate 

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

  x  coordinate 

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

  y  coordinate 

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

  x  coordinate 

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

  y  coordinate 

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

  x  coordinate 

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

  y  coordinate 

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

  x  coordinate 

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

  y  coordinate 

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕