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.
To find the optimal present value of the building, you need to find the value of t0 that gives a point on the graph where the slope of the tangent line is 0.
The first step is to graph the function p(t) = 300,000e^(-0.09t) + (t)/2. You can use any graphing utility or program, such as a graphing calculator or online graphing tool, to plot the function.
After plotting the graph, you can visually examine it to find the point where the slope of the tangent line is 0. This point represents the optimal present value of the building.
Alternatively, you can find the derivative of the function p(t) to determine where the slope of the tangent line equals 0. Let's calculate the derivative:
p'(t) = -0.09 * 300,000e^(-0.09t) + 1/2
To find the value of t where p'(t) = 0, set the derivative equal to 0 and solve for t:
0 = -0.09 * 300,000e^(-0.09t) + 1/2
Simplifying this equation, we get:
0.09 * 300,000e^(-0.09t) = 1/2
Dividing both sides by 0.09 * 300,000 gives:
e^(-0.09t) = 1/0.09 * 300,000
Taking the natural logarithm (ln) of both sides, we obtain:
-ln(0.09t) = ln(1/0.09 * 300,000)
Simplifying further, we get:
-0.09t = ln(1/0.09 * 300,000)
Lastly, solve for t:
t = -(ln(1/0.09 * 300,000))/0.09
Now, you can substitute the value of t into the function p(t) to find the optimal present value of the building.