A craftsman can sell 20 jewelry sets for $600 each. He knows that for each additional set he makes, the price of each set will decrease by $10. How many jewelry sets should he make if he wants to maximize his earnings?
+130544 Let x be the additional number of sets sold after 20....and total revenue = quantity * price
Let the total revenue be represented by R(x)
Let the total sets = (20 + x)
Let the price = (600 - 10x)
So we have
R(x) = (20 + x)(600 - 10x) simplify
R(x) = -10x^2 + 400x + 12000
The number of additional sets sold after 20 is given by the x coordinate of the vertex = -b/ 2a =
-400/ [2(-10)] = -400/-20] = 20
So the number of additional sets sold after 20 = 20 more = 40
And the number of total sets that will maximize revenue = (20 + x) = (20 + 20 ) = 40
And the price that maximizes revenue = (600 - 10(20 ) = (600 - 200) = $400
And the max revenue = (20 + 20)(600 - 10(20) = (40)(600 - 200) = (40)(400) = $16000
