a calculus word problem?

Question

Administrators at a hospital believe that the hospital's expenditures E(B), measured in dollars, are a function of how many beds B are in use with E(B)=14000+(B+1)^2. On the other hand, the number of beds, B, is a function of time t, measured in days, and it is estimated that B(t)=20 sin(t/10) +50.

At what rate, in dollars/day, are the expenditures decreasing when t=100?

Answer 1

E(t)= 14000 + (20 sin(t/10) + 50+ 1)^2
E(t) = 14000 + (20 sin (t/10) + 51)^2
dE/dt (rate of expenditure) = 0 + 2(20 sin(t/10) ) (1/10) (cos(t/10))
dE/dt = 4 sin(t/10) cos(t/10)
at t=100,
dE/dt = 4 sin (10) cos (10)
dE/dt = 4 * (- 0.54402) * (- 0.83907)
dE/dt = 1.826

Answer 2

Substituting for B in the first equation gives you E(t):


E(t) = 14000+[20sin(t/10) + 51]^2

Take the derivative of this wrt t to get

E'(t) = 2*[20sin(t/10) + 51]*(20/10)cos(t/10)

Evaluate at t = 100 to get the answer.

Answer 3

enable x = distance of base of ladder from residing house. enable y = height of authentic of ladder. x² + y² = 29² = 841 even as base of ladder is 21 feet from residing house: y² = 841 - 21² y² = four hundred y = 20 x² + y² = 841 Differentiate each and each and every area with observe of to t 2x dx/dt + 2y dy/dt = 0 y dy/dt = -x dx/dt even as base is 21 feet from the residing house, x = 21, y = 20 Base of ladder is pulled remote from the residing house at a cost of four feet/sec: dx/dt = 4 y dy/dt = -x dx/dt 20 dy/dt = -21 * 4 dy/dt = -80 4/20 dy/dt = -4.2 So authentic of ladder is transferring down the wall at a cost of four.2 feet/sec -------------------- observe: dy/dt = -4.2 The adverse signal exhibits that authentic of ladder is transferring in downward route. So we are saying that ladder is transferring at a cost of -4.2 feet/sec OR ladder is transferring DOWN at a cost of four.2/sec

Answer 4

Put B(t) into E(B) and find the derivative of E(B) with respect to t.
Then put t=100 in the derivative to get your answer.