A sports store orders standard helmets for 20 dollars each and premium helmets for 50 doll… — SAT

A sports store orders standard helmets for $20$ dollars each and premium helmets for $50$ dollars each. The store can spend at most $2{}200$ dollars on this order. The distributor requires the store to order at least $60$ helmets in total. If the store decides to order at least $15$ premium helmets, what is the maximum number of standard helmets that the store can order?

Answer: 72 helmets

Answer

The maximum number of standard helmets the store can order is 72.

By setting the number of premium helmets to its minimum allowed value of

15

to maximize the budget remaining for standard helmets, we find

20x \leq 1450

, which yields

x \leq 72.5

. Since standard helmets must be ordered in whole numbers, the maximum number is

72

, which also satisfies the minimum total order of

60

helmets (

72 + 15 = 87 \geq 60

Step-by-Step Solution

Define variables and translate constraints into inequalities

Let

x

be the number of standard helmets and

y

be the number of premium helmets. The constraints are

20x + 50y \leq 2200

x + y \geq 60

, and

y \geq 15

To represent the problem's mathematical relationships using a system of linear inequalities.

Solve for the upper limit of

x

using the budget constraint and the minimum value of

y

Since

20x \leq 2200 - 50y

x

is maximized when

y

is at its minimum value,

y = 15

. Substituting

y = 15

yields

20x + 750 \leq 2200 \implies 20x \leq 1450 \implies x \leq 72.5

To find the maximum possible value of standard helmets under the budget constraint.

Verify with the minimum total order constraint and determine the maximum integer value

Checking

x + y \geq 60

with

y = 15

gives

x + 15 \geq 60 \implies x \geq 45

. Since

x

must satisfy

45 \leq x \leq 72.5

and must be an integer, the maximum integer value is

72

To ensure the solution is physically possible as a whole number of items and satisfies all constraints.

Key Concept

Systems of Linear Inequalities in Two Variables

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Step-by-Step Solution

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