Quick Computing Company produces calculators They have found
Solution
Let c(x) = ax2 + bx + p, where a, b, c are constants. Now, we know that when x = 2, c (x) = 45, when x = 4, c (x) = 143 and when x = 10, c(x) = 869. Therefore, on substituting these values of x and c(x) in the equation, we have the following equations:
4a + 2b + p = 45...(1)
16a + 4b + p = 143...(2) and
100a + 10b + p = 869...(3)
On subtracting the 1st equation from the 2nd and the 2nd equation from the 3rd equation, we get 12a + 2b = 98..(4) and 84a + 6b = 726..(5). On multiplying the 4th equation by 7 , we get 84a + 14b = = 686...(6). Now, on subtracing the 5th equation from the 6th equation, we get 8b = -40 or, b = -5. Then from the 3rd equation, we have 12a = 2 (-5) = 98 or, 12a = 98 + 10 = 108. Therefore, a = 9. Now from the 1st equation, we have 4*9 + 2(-5) + p = 45 . Therefore, p = 45 -36 + 10 = 19. Thus we have c(x) = 9x2 - 5x + 19. On substituting x = 7 in this equation, we get the costof producing 7 calculators = 9 (7)2 - 5*7 + 19 = 9*49 - 35 + 19 = 441 - 35 + 19 = $425.
