A delivery driver is loading a van with small boxes and large boxes. Each small box weighs… — SAT

A delivery driver is loading a van with small boxes and large boxes. Each small box weighs $10$ pounds, and each large box weighs $35$ pounds. The total weight of the boxes in the van cannot exceed $1,000$ pounds. The driver must load at least $4$ times as many small boxes as large boxes. If the driver loads at least $10$ large boxes, what is the maximum number of large boxes that the driver can load?

Cevap: 13

Cevap

To find the maximum number of large boxes,

y

, we set up the system of inequalities:

10x + 35y \le 1000

and

x \ge 4y

. To maximize

y

, we minimize

x

by setting

x = 4y

. Substituting this into the weight constraint gives

10(4y) + 35y \le 1000

, which simplifies to

75y \le 1000

, or

y \le 13.33

. The largest integer satisfying this inequality is 13, which also satisfies

y \ge 10

Adım Adım Çözüm

Define variables for the quantities of each type of box.

Let

x

be the number of small boxes and

y

be the number of large boxes, where

x

and

y

are non-negative integers.

Establishing variables is necessary to translate the verbal constraints into algebraic inequalities.

Translate the given constraints into a system of linear inequalities.

The weight limit gives

10x + 35y \le 1000

. The requirement for at least 4 times as many small boxes as large boxes gives

x \ge 4y

. The minimum of 10 large boxes gives

y \ge 10

Representing the scenario mathematically allows us to find the boundaries of the solution set.

Express the weight inequality in terms of a single variable by using the boundary condition of the second inequality.

To maximize

y

, we want to minimize the weight contributed by the small boxes,

x

. The minimum value of

x

for any

y

x = 4y

. Substituting

x = 4y

into the weight inequality gives

10(4y) + 35y \le 1000

Finding the extreme case (minimum number of small boxes) provides the upper limit for the number of large boxes.

Simplify the inequality and solve for

y

40y + 35y \le 1000 \implies 75y \le 1000 \implies y \le \frac{1000}{75} \approx 13.33

This calculation determines the upper algebraic bound for the number of large boxes.

Determine the maximum integer value for

y

that satisfies the system.

Since the number of boxes must be an integer, the maximum integer value less than or equal to

13.33

13

. Since

13 \ge 10

, this satisfies all constraints.

Real-world quantities like boxes must be whole numbers, so we take the largest integer within the solution set.

Anahtar Kavram

Solving systems of linear inequalities in two variables to optimize a value under constraints.

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