A studio sells photographs and prints. it costs $50 to purchase each photograph and it take 4 hours to frame it, It costs $100 to purchase each print and it tak

Question

A studio sells photographs and prints. it costs $50 to purchase each photograph and it take 4 hours to frame it, It costs $100 to purchase each print and it take 5 hours to frame it. The store has at most $1800 to spend it and at most 120 hours to frame. It makes $60 profit on each photo. and $100 profit on each print. find the number of each that the studio should purchase to maximize profits.

Answer 1

(This is more than I normally talk all at once in a session. There I be guiding the student to coming up with the equations and interpretations, so take this answer as thought process I will train the student to have)

We need to work out the number of prints and photographs to run through the business. Both words start with the same letter, so I'll say H is the number of photos and R is the number of prints. We're also not necessarily using up all of our time and starting money. We'll need to do those as inequalities to bound where the possible combinations are and then find the highest profit inside that area.

First we'll set up the cost equation. The problem states that photos are $50, prints are $100, and we can't exceed $1800. In mathematical form, that's $1800>=$50*H+$100*R

We're also limited by time. We can't exceed 120 hours, photos take 4 hours, and prints take 5. That will give us a similar statement as before: 120>=4*H+5*R

The last equation we need is the profit equation. Based on the description, profit P is P=$60*H+$100*R.

Now we'll rearrange all of them to solve for a single variable. I picked H, but either would work. Doing that gives us:

H<=36-2R

H<=30-1.25R

H=P/60-(5/3)R

Looking at them, we can see that the slope of the profit line is between the two making up the limits of cost and time. -1.33 is between -1.25 and -2. That means the maximum profit is where the limiting lines intersect. To find that point, we'll set them equal to solve for R and then substitute that back into one of them to get H.

36-2R=30-1.25R (subtract 30 and add 2R to both sides)

6=0.75R (divide both sides by 0.75)

8=R

Plugging that 8 into the first inequality gives

H=36-2(8)=36-16

H=20

If these numbers hadn't come out to be whole numbers, we'd take the nearest set of whole numbers that satisfies both inequalities. Our answer is whole numbers though, so we'll can calculate profit with them.

P=$60(20)+$100(8)=$1200+$800

P=$2000