is given by T(t) = -.1t^2+1.2t+98; for 0 less than or equal to t, less than or equal to 12. Where T is tempt at time t, in days, find the mad value of the temp and when it occurs
what?
2007-07-16 09:01:38 · 3 answers · asked by alfreddood 1 in Science & Mathematics ➔ Mathematics
Find the derivative and set it equal to 0 to find the critical points:
T'(t) = -0.2t + 1.2
-0.2t + 1.2 = 0
-0.2t = -1.2
t = 6
Check for absolute max by plugging the critical points and bounds into the function:
T(6): -0.1(6^2) + 1.2(6) + 98 = 101.6
T(0): -0.1(0^2) + 1.2(0) + 98 = 98
T(12): -0.1(12^2) + 1.2(12) + 98 = 98
Max. is 101.6 and occurs at 6 days.
2007-07-16 09:07:57 · answer #1 · answered by whitesox09 7 · 0⤊ 0⤋
Hey, Alf, your limits are screwed up. They are
t >= 0 and t <= 12. Take the derivitive of T(t), which is -0.2 t + 1.2. Solve for t. Substitute value in equation and solve for T.
2007-07-16 16:08:09 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
Maybe you should clarify your question? The data you have given is absolutely insane (don't get me wrong, I'm not putting you down or anything)
2007-07-16 16:10:06 · answer #3 · answered by Anonymous · 0⤊ 0⤋