hw15 - 6.5 15 p x = 01 x 3 A = 200 So in the notation of...

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Unformatted text preview: § 6.5 15. p ( x ) = . 01 x + 3; A = 200. So in the notation of Figure 8 in the book, the selling price is B = p ( A ) = p (200). The producers’ surplus is integraldisplay 200 B − p ( x ) dx = integraldisplay 200 5 − . 01 x − 3 dx = integraldisplay 200 2 − . 01 x dx = 2 x − . 01 2 x 2 vextendsingle vextendsingle vextendsingle 200 = 400 − . 01 2 (200 2 ) = 400 − 200 = 200 dollars. 16. p ( x ) = x 2 / 9 + 1; A = 3. Producers’ surplus is integraldisplay 3 p (3) − p ( x ) dx = integraldisplay 3 2 − x 2 9 − 1 dx = integraldisplay 3 1 − x 2 9 dx = x − x 3 27 vextendsingle vextendsingle vextendsingle 3 = 3 − 3 3 27 = 2 dollars. 17. p ( x ) = x/ 2 + 7; A = 10. integraldisplay 10 p (10) − p ( x ) dx = integraldisplay 10 12 − x/ 2 − 7 dx = integraldisplay 10 5 − x/ 2 dx = 5 x − x 2 / 4 vextendsingle vextendsingle vextendsingle 10 1 = 50 − 10 2 / 4 = 25 . 18. p ( x ) = 1 + 1 2 √ x ; A = 36. integraldisplay 36 p (36) − p ( x ) dx = integraldisplay 36 4 − 1 − 1 2 √ x dx = integraldisplay 36 3 − 1 2 √ x dx = 3 x − ( 1 2 )( 2 3 ) x 3 / 2 vextendsingle vextendsingle vextendsingle 36 = 3(36) − 1 3 (36 3 / 2 ) = 108 − 1 3 6 3 = 108 − 72 = 36 . 19. Demand: p D ( x ) = 12 − x/ 50; Supply: p S ( x ) = x/ 20 + 5. Optimal production is when p D ( x ) = p S ( x ) 12 − x/ 50 = x/ 20 + 5 7 = x/ 50 + x/ 20 7 = 7 x/ 100 x = 100 . Moreover, p D (100) = 12 − 2 = 10 (and this agrees with p S (100) = 10). So the point of intersection of the curves is ( A, B ) = (100 , 10). Consumers’ surplus: integraldisplay 100 p D ( x ) − p D (100) dx = integraldisplay 100 12 − x/ 50 − 10 dx = integraldisplay 100 2 − x/ 50 dx = 2 x − x 2 / 100 vextendsingle vextendsingle vextendsingle 100 = 2(100) − 100 2 / 100 = 100 dollars. Producers’ surplus: integraldisplay 100 p S (100) − p S ( x ) dx = integraldisplay 100 10 − x/ 20 − 5 dx = integraldisplay 100 5 − x/ 20 dx 2 = 5 x − x 2 / 40 vextendsingle vextendsingle vextendsingle 100 = 5(100) − 100 2 / 40 = 500 − 100( 100 40 ) = 500 − 250 = 250 dollars. 20. Demand: p D ( x ) = √ 25 − . 1 x . Supply: p S ( x ) = √ . 1 x + 9 − 2. Solving for the in- tersection point takes some work here, and one must be careful with the logic, as we discuss below. Curves intersect where p D ( x ) = p S ( x ) √ 25 − . 1 x = √ . 1 x + 9 − 2 We can square both sides and the equality still holds: 25 − . 1 x = ( . 1 x + 9) + 4 + 2( − 2 √ . 1 x + 9) Now we’ll isolate the square root term: 12 − . 2 x = − 4 √ . 1 x + 9 Now square both sides again: 144 + . 04 x 2 − 24( . 2 x ) = 16( . 1 x + 9) Now simplify into standard quadratic form: 144 −...
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This note was uploaded on 04/03/2008 for the course MATH 16A taught by Professor Stankova during the Spring '07 term at Berkeley.

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hw15 - 6.5 15 p x = 01 x 3 A = 200 So in the notation of...

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