The equation to maximize profit is: 2x + y = 500
The equation to satisfy the dealer's demand is: x ≥ 100
The equation to satisfy the ratio of low grade to high grade shoes is: y ≥ 2x
The solution to this system of equations is x = 100 and y = 400. Therefore, the factory should produce 100 pairs of high grade shoes and 400 pairs of low grade shoes for maximum profit.
Suppose ashos factory produces both low grade and high grade shoes.the factory produces at least twice as many low grade as high grade shoes .the maximum possible production is 500 pairs of shoes.adealer calls for delivery of at least 100 high grade pairs of shoes per day.suppose the operation makes aprofit of birr 2 per apair of shoes on high grade shoes and birr 1per apair of shoes on low grade shoes.how many pairs of shoes of each type of should be produced for maximum profit ?
Hint let x denotes the number of high grade shoes and y the number of low grade shoes
2 answers
restraints: x ≥ 100
y ≥ 2x
x + y ≤ 500
profit = 2x + y
sketching this and solving y ≥ 2x with, x + y ≤ 500
I get x = 166, y = 332 , (actually 166 2/3, but we can't make a partial pair
of shoes, and if I go to x = 167 and y = 334, I run contrary to the constrains of x + y ≤ 500
so stay with x = 166, y = 332
this satisfies, y > 2x
and x+y < 500
the max profit would be 2(166) + 332 = 664 birr
(looks the automated answering robot can't handle this type of question)
y ≥ 2x
x + y ≤ 500
profit = 2x + y
sketching this and solving y ≥ 2x with, x + y ≤ 500
I get x = 166, y = 332 , (actually 166 2/3, but we can't make a partial pair
of shoes, and if I go to x = 167 and y = 334, I run contrary to the constrains of x + y ≤ 500
so stay with x = 166, y = 332
this satisfies, y > 2x
and x+y < 500
the max profit would be 2(166) + 332 = 664 birr
(looks the automated answering robot can't handle this type of question)