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.
Answer:
Dh/dt = 0.082 ft/min
Step-by-step explanation:
As a perpendicular cross section of the trough is in the shape of an isosceles triangle the trough has a circular cone shape wit base of 1 feet and height h = 2 feet.
The volume of a circular cone is:
V(c) = 1/3 * π*r²*h
Then differentiating on both sides of the equation we get:
DV(c)/dt = 1/3* π*r² * Dh/dt (1)
We know that DV(c) / dt is 1 ft³ / 5 min or 1/5 ft³/min
and we are were asked how fast is the water rising when the water is 1/2 foot deep. We need to know what is the value of r at that moment
By proportion we know
r/h ( at the top of the cone 0,5/ 2) is equal to r/0.5 when water is 1/2 foot deep
Then r/h = 0,5/2 = r/0.5
r = (0,5)*( 0.5) / 2 ⇒ r = 0,125 ft
Then in equation (1) we got
(1/5) / 1/3* π*r² = Dh/dt
Dh/dt = 1/ 5*0.01635
Dh/dt = 0.082 ft/min
Answer:
120 degrees
Step-by-step explanation:
Angle 1 is congruent to angle 5 by the corresponding angles theorem.