Answer:
Explanation:
The correct question is attached in the image below;
From there, given that: the earnings = 60x + 50 y ------ (1)
y = -x + 175
for x = 0 ⇒ y = 175 ; Then Point A (0, 175)
y = -2x + 300
for y = 0 ⇒ x = 150 ; Then Point C (150, 0)
For Point B;
y = -x + 175 ---- (a)
y = -2x + 300 ---- (b)
Equating both (a) and (b) from above together; then:
⇒ -x + 175 = -2x + 300
⇒ x = 125
∴
From y = -x + 175
y = -125 + 175
y = 50
So point B ( 125, 50)
Now, the points are O(0, 0), A(0, 175), B(125, 50), C(150, 0)
As such, profit at these points are:
O(0, 0), Profit = 60 × 0 + 50 × 0 = 0
A(0, 175), Profit = 60 × 0 + 50 × 175 = 8750
B(125, 5), Profit = 60 × 125 + 50 × 50 = 10,000
C(150, 0), Profit = 60 × 150 + 50 × 0 = 9000
Hence, maximum profit takes place at B(125, 50) which is = $10,000
Finally, we can conclude that the shop should prepare 125 packages of assortment A and 50 packages of assortment B to maximize profit.
The maximum profit is $10,000