The present value of a building in the downtown area is given by the function
p(t) = 300,000e^-0.09t+(t)/2 f or 0 _< t _< 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.
p ' (t) = 300000(-.09) e^(-.09t) + .5
= 0 for a max/min of p(x)
27000 e^(-.09t) = .5
ln both sides
ln 27000 - .09t(lne) = ln .5
ln27000 - ln .5 = .09t , since lne = 1
10.89674 = .09t
t = 121
when t = 121
p(121) = 300000e^-10.89 + 121/2
= 5.593
This result and the original equation makes no sense to me.
Was the t/2 part of the exponent?
If so, then brackets around the exponent would have been essential.
Hello Reiny it is essential
To find the optimal present value of the building, we need to find the value of t that gives a point on the graph (t0, P(t0)), where the slope of the tangent line is 0.
Step 1: Graph the function P(t)
You can use a graphing utility or software to plot the function P(t) = 300,000e^(-0.09t) + (t)/2 for the given values of t, where 0 ≤ t ≤ 10.
Step 2: Locate the point of tangency
Examine the graph and locate the point where the slope of the tangent line is zero. This point will be the optimal value of t that gives the maximum present value of the building.
Step 3: Interpret the results
Identify the coordinates of the point of tangency (t0, P(t0)). The x-coordinate, t0, represents the optimal value of t, and the y-coordinate, P(t0), represents the maximum present value of the building at that time.
Note: The graphing utility or software should provide tools to help you find this point accurately. Confirm your findings by checking for a zero slope or near-zero slope at that point.