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A new company is growing so that its value t years from now will be 50t^2 dollars. Therefore, its present value (at the rate of 8% compounded continuously) is V(t) = (50t^2)(e^-0.08t) dollars (for t >0) Find the number of years that maximizes the present value. Verify answer.
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set derivative equal to 0.....use product rule \[V'(t) = 100t(e^{-.08t}) + 50t^{2}(-.08)e^{-.08t} = 0\]
ok. Thank you.
What is the next step to solve this problem?
factor out the e^-.08t solve resulting quadratic for "t"
notice \[e^{-.08t} = 0\] has no solution
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so at this point the problem would look like.. (e^-0.08t) (100t+(50t^2)(-0.08)) ?
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