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# Chemco

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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.

Feb 11, 2021

### 4+0 Answers

#1
+1

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.

Feb 11, 2021
#2
+1

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)

Feb 11, 2021
#3
+1

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)

Feb 11, 2021
#4
+1

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   Feb 11, 2021