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A deli owner has determined that his revenue, [tex]y[/tex], from selling sandwiches each day is at most [tex]-0.05x^2 + 6x[/tex], where [tex]x[/tex] represents the number of sandwiches he sells. To make a profit, his revenue must be greater than his costs, represented by the expression [tex]1.5x + 45[/tex].

Write a system of inequalities to represent the values of [tex]x[/tex] and [tex]y[/tex] where the deli owner makes a profit. Then complete the statements.

The point [tex](30, 90)[/tex] is [tex]$\square$[/tex] of this system.
The point [tex](60, 160)[/tex] is [tex]$\square$[/tex] of this system.


Sagot :

To write a system of inequalities to represent the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] where the deli owner makes a profit, we need to consider the following conditions:

1. The revenue [tex]\( y \)[/tex] from selling [tex]\( x \)[/tex] sandwiches each day is at most [tex]\( -0.05x^2 + 6x \)[/tex].
2. In order to make a profit, the revenue [tex]\( y \)[/tex] must be greater than the costs which are represented by the expression [tex]\( 1.5x + 45 \)[/tex].

Thus, the system of inequalities can be written as:
[tex]\[ \begin{cases} y \leq -0.05x^2 + 6x \\ y > 1.5x + 45 \end{cases} \][/tex]

Now, let’s complete the statements about the points [tex]\( (30, 90) \)[/tex] and [tex]\( (60, 160) \)[/tex]:

1. For the point [tex]\( (30, 90) \)[/tex]:
- Substituting [tex]\( x = 30 \)[/tex] into the revenue function [tex]\( y \)[/tex], we get [tex]\( y = -0.05(30)^2 + 6(30) = 135 \)[/tex].
- Substituting [tex]\( x = 30 \)[/tex] into the cost function [tex]\( y \)[/tex], we get [tex]\( y = 1.5(30) + 45 = 90 \)[/tex].
- Checking the inequalities: The point [tex]\( (30, 90) \)[/tex] does not satisfy the condition [tex]\( y \leq -0.05(30)^2 + 6(30) \)[/tex] since [tex]\( 90 \leq 135 \)[/tex] is true, but it does not satisfy the profit condition [tex]\( y > 1.5(30) + 45 \)[/tex] since [tex]\( 90 > 90 \)[/tex] is false.

Therefore, the point [tex]\( (30, 90) \)[/tex] is not a solution of this system.

2. For the point [tex]\( (60, 160) \)[/tex]:
- Substituting [tex]\( x = 60 \)[/tex] into the revenue function [tex]\( y \)[/tex], we get [tex]\( y = -0.05(60)^2 + 6(60) = 180 \)[/tex].
- Substituting [tex]\( x = 60 \)[/tex] into the cost function [tex]\( y \)[/tex], we get [tex]\( y = 1.5(60) + 45 = 135 \)[/tex].
- Checking the inequalities: The point [tex]\( (60, 160) \)[/tex] does not satisfy the condition [tex]\( y \leq -0.05(60)^2 + 6(60) \)[/tex] since [tex]\( 160 \leq 180 \)[/tex] is true, but it does not satisfy the profit condition [tex]\( y > 1.5(60) + 45 \)[/tex] since [tex]\( 160 > 135 \)[/tex] is false.

Therefore, the point [tex]\( (60, 160) \)[/tex] is not a solution of this system.

Summary:

- The point [tex]\( (30, 90) \)[/tex] is not a solution of this system.
- The point [tex]\( (60, 160) \)[/tex] is not a solution of this system.