Question 152913
Let E = the monthly cost at the Exton plant to produce x calculators.
Let W = the monthly cost at the Whyton plant to produce (1500-x) calculators.
As you can see, the number of calculators produced by the two plants together is (x+ (1500-x) = 1500).
Let's write the equations for the monthly cost at each plant:
E = $7000+($7.50)(x) This is the cost at the Exton plant.
W = $8800+($6.00)(x) This is the cost at the Whyton plant.
The problem asks "how many calculators should be produced at each plant if the total cost is to be the same at each plant.
So we set E = W and solve for x and then 1500-x.
$7000+$7.50(x) = $8800+$6.00(1500-x) Simplify and solve for x.
7000+7.5x = 8800+9000-6x
7000+7.5x = 17800-6x Subtract 7000 from both sides.
7.5x = 10800-6x  Add 6x to both sides.
13.5x = 10800 Finally, divide both sides by 13.5
x = 800 and (1500-x) = (1500-800) = 700
So the Exton plant must produce {{{highlight(800)}}} calculators and the Whyton plant must produce {{{highlight(700)}}} calculators for a total of 1500 calculators for the total cost at each plant to be the same.
Check:
E = $7000+$7.50(800) = $7000+$6000 = $1300
W = $8800 +$6.00(700) = $8800+$4200 = $1300
The cost is the same at each plant!