### Video Transcript

A small company dyes shirts to be
either solid-color or tie-dye, and they want to decide how many shirts of each color
to prepare for an upcoming sale. The company has a budget of 240
dollars. Purchasing an undyed shirt costs
two dollars. It costs an additional 50 cents to
dye a shirt with a solid color and one dollar 50 to dye a shirt with a tie-dye
pattern. The company only has eight hours to
prepare all the shirts for the sale. And it takes two minutes to produce
a solid-color shirt and 10 minutes to produce a tie-dye shirt. They decide to make a graph to help
them maximize profit, given that they make eight dollars profit for each solid-color
shirt and 10 dollars profit for each tie-dye shirt. Let π₯ represent the number of
solid-color shirts and π¦ represent the number of tie-dye shirts.

There are three parts to this
question, which we will look at shortly. Before doing this, letβs consider
some of the information we are given. There are two types of shirts made
by the company, solid-color shirts represented by π₯ and tie-dye shirts represented
by π¦. We will solve this problem using
linear programming by firstly defining an objective function and some constraints in
the form of linear inequalities. The company is looking to maximize
profit. And we are told they make an
eight-dollar profit for each solid-color shirt and a 10-dollar profit for each
tie-dye shirt. This means that the profit in
dollars π is equal to eight π₯ plus 10π¦. We can also write this as a
function in terms of π₯ and π¦ such that π of π₯, π¦ is equal to eight π₯ plus
10π¦.

The constraints placed on the
company involve both time and money. If we consider money first, we are
told that the company has a budget of 240 dollars. We are told that an undyed shirt
costs two dollars and that it costs an additional 50 cents to dye a shirt with a
solid color and one dollar 50 to dye a shirt with a tie-dye pattern. It therefore costs two dollars 50
to make each solid-color shirt. As the company are making π₯ of
these, this can be written as 2.5π₯. It costs three dollars 50 to make a
tie-dye shirt. And this can be written as
3.5π¦. We know that the sum of these terms
must be less than or equal to 240 as the total budget was 240 dollars.

Letβs now consider the time
constraints. We are told it takes two minutes to
produce a solid-color shirt and 10 minutes to produce a tie-dye shirt. The company has eight hours or 480
minutes to prepare all the shirts. This can be represented by the
inequality two π₯ plus 10π¦ is less than or equal to 480. We will now clear some space and
consider the three parts of this question.

The three parts to the question are
as follows. Which of the following shows the
feasible region? State the objective function. How many of each type of shirt
should the company produce to maximize profit?

We have already answered the second
part of this question. The objective function or profit
function in this question π of π₯, π¦ is equal to eight π₯ plus 10π¦. In the first part of this question,
weβre given four graphs with the straight lines 2.5π₯ plus 3.5π¦ equals 240 and two
π₯ plus 10π¦ equals 480 drawn on them. We know that two π₯ plus 10π¦ must
be less than or equal to 480. This means that the feasible region
lies below this line. We can therefore rule out option
(B).

We are also told that 2.5π₯ plus
3.5π¦ is less than or equal to 240. The feasible region must therefore
also lie below the blue line on our graphs. This rules out option (C). To decide whether graph (A) or
graph (D) is the correct one, we need to consider two further constraints. Since we canβt make a negative
number of shirts, both π₯ and π¦ must be greater than or equal to zero. This means that the feasible region
must lie above the π₯-axis and to the right of the π¦-axis. We can therefore rule out option
(A) as part of the feasible region here occurs when π₯ is less than zero. The correct answer to the first
part of our question is therefore option (D).

We will now clear some space and
solve the third part of this question. We recall that this asked us to
work out the number of each type of shirt the company should produce to maximize
profit. We recall that the profit or
objective function was equal to eight π₯ plus 10π¦ and that the feasible region was
subject to four constraints. We know that the possible maximum
values of the function occur at the vertices of the feasible region. So we will begin by working out the
coordinates of these points. Firstly, we have the point zero,
zero. We know that when a line intersects
the π₯-axis, π¦ is equal to zero. And substituting this into the
equation, 2.5π₯ plus 3.5π¦ equals 240, we find that π₯ is 96. So the coordinates of this vertex
are 96, zero.

In the same way, when a line
intersects the π¦-axis, our π₯-coordinate is zero. Substituting this into the equation
two π₯ plus 10π¦ equals 480 gives us π¦ is equal to 48 and another vertex at zero,
48. The fourth vertex of our feasible
region is the point of intersection of our two diagonal lines. One way of working out this point
would be to find a solution of the simultaneous equations shown. There are many ways of solving
these. One way would be to multiply the
second equation by two. This gives us the equation five π₯
plus seven π¦ is equal to 480.

Rewriting the first equation
underneath and then subtracting the two equations, we have three π₯ minus three π¦
is equal to zero. Adding three π¦ to both sides of
this equation and then dividing through by three, we see that π₯ is equal to π¦. We can then substitute this back
into our first equation such that two π¦ plus 10π¦ is equal to 480. Solving this gives us π¦ is equal
to 40 and as π₯ is also equal to this. The fourth vertex has coordinates
40, 40.

We can now substitute each of these
coordinates in turn into our objective function. It is important to note here that
any of these points could be the optimal solution and it will not necessarily be the
point of intersection we have just found. Substituting in the values of π₯
and π¦, the objective function is equal to zero, 480, 768, and 720. As the company is trying to
maximize the profit, they choose the highest value. Since π₯ represents the number of
solid-color shirts and π¦ the number of tie-dye shirts, producing 96 solid-color
shirts and zero tie-dye shirts would maximize the profit. Assuming that all the shirts were
sold, this would yield a profit of 768 dollars.