Answer:
$3
Step-by-step explanation:
Given that:
p = 8 - ln(x) when 5 < x < 500
where;
x = The total number of dogs sold
Then;
The total revenue = x * p
R = x(8 - ln(x))
R = 8x - xln(x)
The Company thus pays 1 dollar per dog
i.e.
The total cost C = 1 * x = x
Then: Profit = R - C
P = 8x - xln(x) - x
P = 7x - xln(x)
Differentiating P in respect to x
dP/dx = 7 - d/dx(xln(x))
dP/dx = 7 - x*d/dx(ln(x)) - ln(x)*d/dx(x)
dP/dx = 7 - x(1/x) - ln(x)
dP/dx = 6 - ln(x)
Since this must be maximized, dP/dx is set to be equal to 0
6 - ln(x) = 0
ln(x) = 6
x = e^6
Now, p = 8 - ln(x)
Plug in the value of x :
p = 8 - ln(e^5)
p = 8 - 5
p = 3
Therefore, each dog must be priced at $3 to maximize the profit.
0.1666666667 cups of water
90 mins = 0.25
60 min = 0.1666666667
102,650-82,400=20,250. The value of the home increased by $20,250.
Answer:
58/9
Step-by-step explanation:
5/3 +43/9
Lets take the L.C.M first
The L.C.M would be 9
Solve the term by taking 9 as L.C.M
=15+43/9
Add the numerator.
=58/9
The answer is 58/9 ....
