Marissa is a photographer. She sells framed photographs for $100 each and greeting cards for $5 each. The materials for each framed photograph cost $30, and the materials for each greeting card cost $2. Marissa can sell up to 8 framed photographs and 40 greeting cards each week, but this week, she only has $200 to spend on materials. Marissa hopes to earn a profit of at lest $400 this week after paying for her materials.

Question

Let x = the number of framed photographs.
Let y = the number of greeting cards.
Two of the inequalities that model this situation are x < =8 and y < = 40

Explain your inequalities and explain why x<=8 and y<=40 are also inequalities for this system

I got Material cost equation : 30x + 2y = 200
Profit equation : 100x + 5y = 400

Answer 1

Aren't they supposed to be in-equalities?

profit = revenue - expenses
= 100x + 5y - 30x - 2y
= 70x + 3y

but
100x + 5y - 30x - 2y ≥ 400
70x + 3y ≥ 400

also 30x + 2y ≤ 200
15x + y ≤ 100

" Marissa can sell up to 8 framed photographs" --- > x ≤ 8
" and 40 greeting cards each week" ---> y ≤ 40

I would also include x≥ 0, y ≥ 0

solving the 2 linear relations as equations:
70x + 3(100-15x) = 400
70x + 300 - 45x = 400
25x = 100
x = 4
y = 40

She should make and sell 4 photographs, and 40 cards

profit+ 70(4) + 3(40) = 400

suppose she goes for 6 photographs and 40 cards ...
x=6 , y = 40
profit > 400
both x and y agree with x≤8 and y≤ 40
BUT
30(6) + 2(40) = $260
and she only wanted to spend a maximum of $200 on materials

Answer 2

Im confused

Answer 3

Marissa plans to make and sell 5 framed photographs and 25 greeting cards. Is that a solution to the system of inequalities? If so, is it the solution that will produce the most profit?