Question 417·Hard·Algebra

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A smartphone manufacturer estimates that the total production cost, $C$ , in dollars, of assembling $x$ phones in a given week can be modeled by a linear function. Data from recent weeks show that producing 400 phones costs $54,000 and producing 1,100 phones costs $99,000. The factory's weekly budget for production is $84,000. According to the model, what is the maximum number of phones the factory can assemble in a week without exceeding its production budget?

A linear model adds the same amount to production cost for each additional phone. Use the two known weeks to find that per-phone increase, then build a cost rule and use a Desmos regression to find where it reaches the budget. The maximum whole-number count is below the budget boundary if the boundary falls between two integers; rounding up would exceed the budget.

Hints

Hint 1
The slope is the added cost for one more phone. Divide the change in cost by the change in phone count. Dividing one week's total cost by its phone count would assume there is no fixed cost.
Hint 2
A known cost belongs to a known phone count, not to zero phones. Start with the cost at $400$ phones, then add the per-phone increase for each phone beyond $400$ .
Hint 3
Without exceeding means the cost may equal the budget but cannot go above it. Once you find the count where the model reaches the budget, which whole-number count is the greatest one still allowed?

Step-by-step

Build the cost model and find the budget boundary
Step 1
Pair each phone count with its cost
The input is phones assembled, and the output is total cost. So the two weeks give two points on the cost line: $C(400)=54000$ and $C(1100)=99000$ .
Step 2
Find the cost per additional phone
The slope is the cost increase divided by the phone increase. Type $\frac{99000-54000}{1100-400}$ in Desmos. It shows about $64.2857$ ; use its fraction button to see $\frac{450}{7}$ dollars per phone.
Step 3
Build the cost rule from a known week
At $400$ phones, cost is $54000$ . Each of the $x-400$ additional phones adds $\frac{450}{7}$ dollars, so start from that known cost: $C(x)=54000+\frac{450}{7}(x-400)$ . Type this function in Desmos for the next step.
Step 4
Turn the budget into a limit
Without exceeding the $84000$ budget means the cost can equal $84000$ or be less: $C(x)\le84000$ . Because the per-phone cost is positive, the largest allowed count is at or below where $C(x)=84000$ .
Step 5
Find where cost reaches the budget
Type $C(x_1)\sim84000$ . The subscript in $x_1$ tells Desmos to solve for a value rather than graph $x$ , and $\sim$ asks it to fit the equality. Under PARAMETERS, it shows the budget boundary at $x_1\approx866.6667$ phones.
Step 6
Choose the last whole number allowed
Phones come in whole numbers. The greatest whole number at or below $866.6667$ is $866$ ; rounding up to $867$ would exceed the budget. So the factory can assemble at most 866 phones in a week. Choice B.

Question 417·Hard·Algebra

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