Optimal Price

Calculus Level pending

The cost of publishing a book is a fixed $1200 (initial cost) plus $9 per copy. At $15 selling price, it is expected the number of copies sold will be 300, and that each increment of the price by $1 will reduce the number of copies sold by 10. What is the optimal selling price of the book (that will maximize the profit of selling the book) ?

$21.50 $27.00 $19.90 $15.00

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2 solutions

Tom Engelsman
Mar 5, 2021

Taking the revenue as Sales - Costs, we can write the following function:

R ( x ) = ( 300 10 x ) ( 15 + x ) [ 9 ( 300 10 x ) + 1200 ] = 10 x 2 + 240 x + 600 = 10 ( x 2 24 x ) + 600 = 10 ( x 12 ) 2 + 2040 R(x) = (300-10x)(15+x) - [9(300-10x) + 1200] = -10x^2 + 240x + 600 = -10(x^2 -24x) + 600 = \boxed{-10(x-12)^2 + 2040}

which attains its maximum value of $ 2 , 040 2,040 at x = 12 x =12 , or 300 10 ( 12 ) = 180 300-10(12) = 180 copies. This yields at optimized selling price of 15 + 12 = 27 15 + 12 = \boxed{27} dollars.

Pop Wong
Mar 5, 2021

Let

  • p p be the selling price
  • C C be the cost
  • x x be the number of copies
  • R R be the revenue

C = 1200 + 9 x x = 300 ( p 15 ) 10 = 450 10 p R = p x C = p x 1200 9 x = ( p 9 ) x 1200 R = ( p 9 ) ( 450 10 p ) 1200 = 10 ( ( p 9 ) ( 45 p ) 120 ) R = 10 ( p 2 + 54 p 525 ) C = 1200 + 9x\\ x = 300 - (p-15)10 = 450 - 10p \\ R = px - C = px - 1200 - 9x = (p-9)x - 1200\\ R = (p-9)(450 - 10p) - 1200 = 10( (p-9)(45-p) - 120 )\\ R = 10( -p^2 + 54p -525 )\\

max R max ( p 2 + 54 p 525 ) \max R \implies \max (-p^2 + 54p -525 )

p 2 + 54 p 525 -p^2 + 54p -525 is concave downward.

Its maximum when p = α + β 2 = 54 2 = 27 p = \cfrac{\alpha + \beta}{2} = \cfrac{54}{2} =\boxed{27}

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