Answer:
€ 270
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