A candy shop puts together two prepackaged assortments to be given to​ trick-or-treaters on Halloween. The feasible set for producing x packages of assortment A and y packages of assortment B has been graphed to the right. For the earnings function Earningsx​y, where Earnings is in​ cents, how many of each assortment should be prepared in order to maximize​ profits? What is the maximum​ profit? 0 600 0 600 x y A coordinate system has a horizontal x-axis labeled from 0 to 600 in increments of 120 and a vertical y-axis labeled from 0 to 600 in increments of 120. A four-sided polygon labeled as the feasible set is formed by a series of line segments and their intersection with the x-axis and y-axis. The line segments are labeled as follows: y equals negative 2 x plus 600; y equals negative x plus 350. The shop should prepare nothing packages of assortment A and nothing packages of assortment B to maximize profit. ​(Type whole​ numbers.) The maximum profit is ​$ nothing. ​(Round to the nearest cent as​ needed.)

Step-by-step 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

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