The present value of a building in the downtown area is given by the function
P(t) = 300,000e^-0.09t+¡Ìt/2 f or 0 ¡Ü t ¡Ü 10
Find the optimal present value of the building. (Hint: Use a graphing utility to graph the function, P(t), and find the value of t0 that gives a point on the graph, (t0, P(t0)), where the slope of the tangent line is 0.)
To find the optimal present value of the building, we need to find the value of t that gives a point on the graph where the slope of the tangent line is 0. This means we are looking for the maximum or minimum point on the graph.
To start, let's graph the function P(t) = 300,000e^(-0.09t+√(t/2)) for the given range 0 ≤ t ≤ 10 using a graphing utility.
Once we have the graph, we need to find the value of t (denoted as t0) where the slope of the tangent line is 0. This indicates the maximum or minimum point on the graph.
Using the graphing utility, locate the point on the graph where the slope of the tangent line is 0. This point is denoted as (t0, P(t0)), where t0 is the value we are looking for.
Now that we have the value of t0, substitute it back into the original equation P(t) = 300,000e^(-0.09t+√(t/2)) to find the optimal present value of the building.