Chemco, a chemical company, estimates the profit in thousands of dollars after producing “x” hundred units. Profit can be expressed as,
P(x) = -54x^2 +162x -108
(a) Find the number of units to obtain the maximum profit.
(b) What is the maximum profit?
(c) Find the least and greatest numbers of units that should be produced in order for the company to make a profit.
+9674 P(x) = -54 (x^2 - 3x + 2)
P(x) = -54 (x^2 - 3x + (3/2)^2 + 2 - 9/4)
P(x) = -54(x - 3/2)^2 + 27/2
Number of units = 1.5 hundred
Maximum profit = 13.5 thousand dollars
Try to do part (c) on your own.
Max occurs at x = -b/2a = - 162 /(2 * -54)= 1.5 hundred units = 1500 units
use this value of x =1.5 in the equation to calculate the profit (in thousands of dollars)
Use quadratic formula to find https://web2.0calc.com/questions/power-meassurement
the zeroes of the equation...between these zeroes.. this is
the portion of the equation above zero....where profit is made ( positive)
+130071 P(x) = -54x^2 +162x -108
(a) The x that gives the max profit is -162 / (2 * -54) = -162 / -108 = 1.5 = 1500 units
(b) Max profit is -54(1.5)^2 + 162 ( 1.5) - 108 = 13.5 = $13,500
(c) See the graph here : https://www.desmos.com/calculator/lsabrvvq2g
The least number of units that should be produced is at (1,0) = 1000 units
The greatest number of units that should be produced is at ( 2,0) = 2000 units
