Worded optimisation problem

Question

I have the following question:

A pet food manufacturer produces two types of food: Regular and Premium. A 20kg bag of regular food requires 3 hours to prepare and 3 hours to cook. A 20kg bag of premium food requires 3 hours to prepare and 6 hours to cook. The materials used to prepare the food are available 7 hours per day, and the oven used to cook the food is available 17 hours per day. The profit on a 20kg bag of regular food is 40 dollars and on a 20kg bag of premium food is $49.

The manager wants xbags of regular food and ybags of premium pet food to be made in a day.

(b) Enter the constraint imposed by available preparation time. It will be an inequality involving x and y

(c) Enter the constraint imposed by available time in the oven.

I thought that the answer to b) would be: 3x+3y>=6, but this is wrong... I got 3x as the question said it takes 3 hrs to prepare regular food, and 3y because it said it takes 3 hours to prepare premium. I got 6 by adding these hours together..

Thank you for your time.

Answer 1

First of all one should define the variables:

$x:=$ Number of $20$kg bags which contain regular food.

$y:=$ Number of $20$kg bags which contain premium food.

Then you have a constraint for the availability of the materials, which are necessary to prepare the food. A $20$kg bag regular food needs $3$ hours to prepare and a $20$kg bag premium food needs $3$ hours to prepare as well. The materials are (at most) $7$ hours per day available. And I assume that the regular and the premium food cannot be prepared at the same time. Then the constraint is

$$3x+3y\leq 7$$

Can you proceed with the constraint for the availability of the oven?

Answer 2

If $20$kg requires $3$ hours ($180$ minutes) to prepare, then $1$kg requires $3/20$ hours ($9$ minutes) to prepare. The definition of $x$ isn't clear, but I suppose is $x$ kg. With this in mind, the constraint in the preparation time is \begin{align*} \frac{3}{20}x + \frac{3}{20}y & \le 7 \end{align*} where $\dfrac{3}{20}$ is the rate to prepare both regular and premium food and less than or equal to 7 hours is because the question said "The materials used to prepare the food are available $7$ hours per day". Let us check that the $\dfrac{3}{20}x$ has the correct unit as 7, which is hours: $$ \frac{3}{20}x = \frac{\textrm{[time]}}{\textrm{[mass]}}\textrm{[mass]} = \frac{\textrm{hours}}{\textrm{kg}}\textrm{kg} = \textrm{hours}. $$