# Technically there is a critical number at t 0 and f 0

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First we find the critical numbers. Technically there is a critical number att= 0, andf(0) = 30, but it is okayif the student forgets the critical number oft= 0. There is much more going on in the problem, and if you testa few points you’ll see thatt= 0 is clearly not going to give the maximum value here anyway.So, we solvef0(t) = 0.f0(t) = 0 = 30(1 + 20t)· -0.05e-0.05t+ 30202te-0.05t,0 = 30e-0.05t(1 + 20t)· -0.05 +10t= 30e-0.05t-0.05-t+10t.Since e to any power is positive, the above equation equals zero when its second factor is zero, which is when (acalculator may be used here)t=8001-160018009.84 years.For this value oftwe would getf(t)\$1169.98, which is quite pricey, far better than the valuef(0) = 30 thathe would get for selling the wine now.Does the nonzero critical number give the maximum value off?Note that the distinguishing term for the sign off0, namely-0.05-t+10t, is adecreasingfunction fort >0 since its derivative is negative (-12t-5t3/2<0 ift >0). This means that-0.05-t+10tis positive ifand only if 0< t <8001-16001800, and is negative ift >8001-16001800.Therefore,f0(t)>0 if 0< t <8001-16001800andf0(t)<0 ift >8001-16001800, hence our critical number abovegives the absolute maximum off.Therefore, the dealer should sell the wine after about 9.84 years.
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