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Sagot :
Let's carefully examine the problem to identify the correct constraints.
Lee washes houses, and the time and water constraints for each type of house are given as follows:
- It takes 40 minutes to power wash a one-story house and uses 180 gallons of water.
- It takes 90 minutes to power wash a two-story house and uses 200 gallons of water.
- Lee's total work time per week does not exceed 40 hours.
- Lee uses no more than 5000 gallons of water per week.
We need to formulate the constraints for the number of one-story houses [tex]\( x \)[/tex] and two-story houses [tex]\( y \)[/tex] that Lee can wash in a week based on these conditions.
First, let's address the time constraint:
- The time taken to wash [tex]\( x \)[/tex] one-story houses is [tex]\( 40x \)[/tex] minutes.
- The time taken to wash [tex]\( y \)[/tex] two-story houses is [tex]\( 90y \)[/tex] minutes.
- Lee has a maximum of [tex]\( 40 \)[/tex] hours per week to work. Converting this to minutes gives [tex]\( 40 \times 60 = 2400 \)[/tex] minutes.
Thus, the time constraint is:
[tex]\[ 40x + 90y \leq 2400 \][/tex]
Next, let's address the water constraint:
- The amount of water used to wash [tex]\( x \)[/tex] one-story houses is [tex]\( 180x \)[/tex] gallons.
- The amount of water used to wash [tex]\( y \)[/tex] two-story houses is [tex]\( 200y \)[/tex] gallons.
- Lee uses a maximum of [tex]\( 5000 \)[/tex] gallons of water per week.
Thus, the water constraint is:
[tex]\[ 180x + 200y \leq 5000 \][/tex]
Lastly, since [tex]\( x \)[/tex] and [tex]\( y \)[/tex] represent the number of houses, they must be non-negative:
[tex]\[ x \geq 0 \][/tex]
[tex]\[ y \geq 0 \][/tex]
In summary, the constraints for the problem are:
[tex]\[ \begin{array}{lcl} 40x + 90y & \leq & 2400 \\ 180x + 200y & \leq & 5000 \\ x & \geq & 0 \\ y & \geq & 0 \\ \end{array} \][/tex]
Thus, the correct set of constraints looks like this:
[tex]\[ \boxed{ \begin{array}{l} 40x + 90y \leq 2400 \\ 180x + 200y \leq 5000 \\ x \geq 0 \\ y \geq 0 \end{array} } \][/tex]
Lee washes houses, and the time and water constraints for each type of house are given as follows:
- It takes 40 minutes to power wash a one-story house and uses 180 gallons of water.
- It takes 90 minutes to power wash a two-story house and uses 200 gallons of water.
- Lee's total work time per week does not exceed 40 hours.
- Lee uses no more than 5000 gallons of water per week.
We need to formulate the constraints for the number of one-story houses [tex]\( x \)[/tex] and two-story houses [tex]\( y \)[/tex] that Lee can wash in a week based on these conditions.
First, let's address the time constraint:
- The time taken to wash [tex]\( x \)[/tex] one-story houses is [tex]\( 40x \)[/tex] minutes.
- The time taken to wash [tex]\( y \)[/tex] two-story houses is [tex]\( 90y \)[/tex] minutes.
- Lee has a maximum of [tex]\( 40 \)[/tex] hours per week to work. Converting this to minutes gives [tex]\( 40 \times 60 = 2400 \)[/tex] minutes.
Thus, the time constraint is:
[tex]\[ 40x + 90y \leq 2400 \][/tex]
Next, let's address the water constraint:
- The amount of water used to wash [tex]\( x \)[/tex] one-story houses is [tex]\( 180x \)[/tex] gallons.
- The amount of water used to wash [tex]\( y \)[/tex] two-story houses is [tex]\( 200y \)[/tex] gallons.
- Lee uses a maximum of [tex]\( 5000 \)[/tex] gallons of water per week.
Thus, the water constraint is:
[tex]\[ 180x + 200y \leq 5000 \][/tex]
Lastly, since [tex]\( x \)[/tex] and [tex]\( y \)[/tex] represent the number of houses, they must be non-negative:
[tex]\[ x \geq 0 \][/tex]
[tex]\[ y \geq 0 \][/tex]
In summary, the constraints for the problem are:
[tex]\[ \begin{array}{lcl} 40x + 90y & \leq & 2400 \\ 180x + 200y & \leq & 5000 \\ x & \geq & 0 \\ y & \geq & 0 \\ \end{array} \][/tex]
Thus, the correct set of constraints looks like this:
[tex]\[ \boxed{ \begin{array}{l} 40x + 90y \leq 2400 \\ 180x + 200y \leq 5000 \\ x \geq 0 \\ y \geq 0 \end{array} } \][/tex]
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