Question 585·Hard·Algebra

Let $x$ be the number of premium boxes and $y$ the number of standard boxes. Together they make $x+y$ boxes. At least $200$ means $200$ or more, so $x+y\ge200$. Both counts must be nonnegative whole numbers.

Each premium box costs $7 and each standard box costs $4, so the total cost is $7x+4y$. The $1,000 budget is a maximum, not an amount the caterer must spend: $7x+4y\le1000$.

Premium boxes contribute $400x$ calories and standard boxes contribute $250y$. No more than $65{,}000$ means $400x+250y\le65000$. A pair of counts must pass this limit as well as the first two.

Type the three inequalities into Desmos, then add $x\ge0$ and $y\ge0$ because box counts can't be negative. Desmos shades the pairs that pass each limit. Their overlap ends on the right where the total-box and budget boundaries meet; the calorie limit doesn't cut off that tip.

For a fixed premium count, use the fewest standard boxes allowed when testing the budget. Subtract $x$ from the total-box requirement: $y\ge200-x$. Every standard box beyond that minimum costs another $4, so adding one cannot help fit the budget.

Substitute the smallest allowed standard count, $200-x$, into the cost limit: $7x+4(200-x)\le1000$. If even this cheapest way to make enough boxes costs too much, no larger value of $y$ can work.

Type $7x_1+4(200-x_1)\sim1000$ in Desmos. Here $x_1$ is the premium count at the budget boundary, and $\sim$ asks Desmos to find it. Under PARAMETERS, Desmos reports $x_1\approx66.6667$. Replacing a standard box with a premium box raises the cost by $3, so larger premium counts cannot fit the budget when at least $200$ boxes are required.

The greatest whole number below the cutoff is $66$. With $66$ premium boxes, make $200-66=134$ standard boxes. Type both totals into Desmos: it shows $998 for cost and $59{,}900$ for calories. The counts add to $200$, and both totals are within their limits. So the greatest possible number of premium snack boxes is $66$. Choice B.