Question 525·Hard·Algebra

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During an 8-hour manufacturing shift (480 minutes), a machine makes two kinds of parts, type A and type B. Producing one part of type A keeps the machine running for 6 minutes, and producing one part of type B keeps it running for 9 minutes.

Let $x$ be the number of type A parts and $y$ be the number of type B parts produced in a shift.

Because the machine cannot run for more than the length of the shift, $6x+9y\le480$ .
Because at least 40 parts must be produced, $x+y\ge40$ .

What is the greatest possible value of $y$ ?

Your answer

A system of inequalities describes combinations that meet every limit. Graph the limits in Desmos, including the fact that part counts can't be negative, and look at the top of the shaded overlap. Desmos includes fractional counts, which aren't possible here. For a greatest possible count, find an upper bound and then check that a whole-number pair reaches it.

Hints

Hint 1
The feasible region is the overlap where every condition holds. Part counts can't be negative, even though the question doesn't list that as an inequality. What boundaries should you add before looking for the highest point?
Hint 2
Each type A part uses time that could be spent making type B parts. To make as many B parts as possible, what is the smallest possible value of $x$ ?
Hint 3
An upper bound is a value the count can't exceed. If Desmos gives a fractional bound, remember that parts come in whole numbers. After rounding down, does your proposed pair also meet the minimum of $40$ parts?

Step-by-step

Graph the limits, then check the highest whole-number count
Step 1
Find the top of the possible region
Type both given inequalities into Desmos. Add $x\ge0$ and $y\ge0$ because you can't make a negative number of parts. Desmos shades the overlap, where every condition holds; its highest edge meets the time-limit line at $x=0$ .
Step 2
Leave the most time for type B
Each type A part uses $6$ minutes, so $x=0$ leaves the most time for type B. Put $x=0$ into the time limit:
$6(0)+9y\le480$
Remove the zero term:
$9y\le480$
You still need to check the $40$ -part minimum; saving time alone doesn't prove a shift works.
Step 3
Find the upper bound
At the boundary, the available $480$ minutes are used exactly. Type $9y_1\sim480$ ; $\sim$ tells Desmos to find the value of $y_1$ that makes the sides equal. It shows $y_1\approx53.333$ under PARAMETERS. That's a fractional upper bound, so the greatest whole-number count must be below it.
Step 4
Check that the whole-number count works
Round the upper bound down, not up, and test $x=0$ , $y=53$ . Type $9(53)$ and $0+53$ ; Desmos shows $477$ minutes and $53$ parts. So $477\le480$ and $53\ge40$ : this pair meets both limits. No greater whole-number value of $y$ fits below the upper bound. The greatest possible number of type B parts is $53$ . Grid in 53.

Question 525·Hard·Algebra

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