A chemical manufacturer sells ammonia at a price of 500 dollars per unit. The daily total production cost in dollars for x units is given by 10000+30*x+1/20*x^2, and the daily production capacity is at most 15,000 units.
How many units of ammonia must be manufactured and sold daily to maximize the profit?
and
How much money will be made each day for that amount of production?
Please explain, and do not round answers. TY!
How many units of ammonia must be manufactured and sold daily to maximize the profit?
and
How much money will be made each day for that amount of production?
Please explain, and do not round answers. TY!
-
if you get the value for the sales and then subtract the costs, you get the profit
sales value is : 500 x (500 dollars times x, with x being the number of units sold)
cost is : 10000+30*x+1/20*x^2
profit then is given by
500x - (10000+30*x+1/20*x^2)
for convenience we can rewrite this as
-1/20x^2 + 500x -30x - 10000
-1/20x^2 + 470x - 10000
if we take the first derivative of this expression with respect to x, and then set equal to zero, we can find maxima and minima, which will give us the number "x" that provides the most profit
if f(x)= -1/2 0x^2 + 470x - 10000
f'(x)=-1/10 x+470=0
-x+4700=0
x=4700
f''(x)= -1/10 , this is negative, meaning we have a maximum
so, making 4700 units per day makes the most profit, more or less production both give less profit
at x=4700, the profit is
-1/20x^2 + 470x - 10000
-1/20(4700)^2 + 470(4700) - 10000 =
(to prevent me from being tempted to rounding, I will let you do the arithmetic)
sales value is : 500 x (500 dollars times x, with x being the number of units sold)
cost is : 10000+30*x+1/20*x^2
profit then is given by
500x - (10000+30*x+1/20*x^2)
for convenience we can rewrite this as
-1/20x^2 + 500x -30x - 10000
-1/20x^2 + 470x - 10000
if we take the first derivative of this expression with respect to x, and then set equal to zero, we can find maxima and minima, which will give us the number "x" that provides the most profit
if f(x)= -1/2 0x^2 + 470x - 10000
f'(x)=-1/10 x+470=0
-x+4700=0
x=4700
f''(x)= -1/10 , this is negative, meaning we have a maximum
so, making 4700 units per day makes the most profit, more or less production both give less profit
at x=4700, the profit is
-1/20x^2 + 470x - 10000
-1/20(4700)^2 + 470(4700) - 10000 =
(to prevent me from being tempted to rounding, I will let you do the arithmetic)