Answer is: option4

135 dollars/day

Solution:

Expenditures as a function of beds: \[ E(B) = 14000 + (B + 1)^2 \] Beds as a function of time: \[ B(t) = 20 \sin \left( \frac{t}{10} \right) + 50 \] Compute derivatives: \[ \frac{dE}{dB} = 2(B + 1) \] \[ \frac{dB}{dt} = 20 \cos \left( \frac{t}{10} \right) \times \frac{1}{10} \] \[ \frac{dB}{dt} = 2 \cos \left( \frac{t}{10} \right) \] Compute \( \frac{dE}{dt} \): \[ \frac{dE}{dt} = \frac{dE}{dB} \times \frac{dB}{dt} \] Substituting the derivatives: \[ \frac{dE}{dt} = 2(B + 1) \times 2 \cos \left( \frac{t}{10} \right) \] \[ \frac{dE}{dt} = 4(B + 1) \cos \left( \frac{t}{10} \right) \] Compute \( B(100) \): \[ B(100) = 20 \sin \left( \frac{100}{10} \right) + 50 \] \[ B(100) = 20 \sin(10) + 50 \]

Using \( \sin(10) \approx -0.544 \) (since 10 radians is approximately 572° and sine is negative in the fourth quadrant):

\[ B(100) = 20(-0.544) + 50 \] \[ B(100) \approx -10.88 + 50 = 39.12 \] Compute \( \cos(10) \): \[ \cos(10) \approx -0.839 \] Compute \( \frac{dE}{dt} \): \[ \frac{dE}{dt} = 4(39.12 + 1)(-0.839) \] \[ \frac{dE}{dt} = 4(40.12)(-0.839) \] \[ = 160.48 \times (-0.839) \] \[ \frac{dE}{dt} \approx -134.6 \] Final Answer:

The rate at which the expenditures are decreasing is approximately 135 dollars per day, which corresponds to:

\[ \boxed{\textbf{135 dollars/day}} \]