Question

q

Answer 1

During the time interval 0 is smaller than or equal to t, that is smaller than or equal to 12 hours, the amount of heating oil in tank is given by \(H(t)=\frac{1}{6}t^3-\frac{11}{3}t^2+20t+125\) gallons, where t is measured in hours. To the nearest whole number, what is the least amount of heating oil in the tank?

Answer 2

it's just this one that i'm stuck on

Answer 3

I have \(H'(t)=\frac{1}{2}t^2-\frac{22}{3}t+20\)

Answer 4

and then I'm supposed to find the critical numbers...but I can't really factor \(H'(t)=3t^2-44t+120\)

Answer 5

@darkknight

Answer 6

by the way, I am solving this using the candidates test (using critical numbers and endpoints)

Answer 7

hmm, so you are right, you can't factor that out, so I'd try to use the quadratic formula, your endpoints are 0 and 12 because of the bounds, and you use the numbers you get from the quadratic formula. Too lazy to do the math so I am going on mathway but it ain't loading on my end

Answer 8

wait wat wait I get 2 negative numbers when I find the roots of this, this is the right equation?

Answer 9

No I plugged the wrong eqn in, 1 sec

Answer 10

Yes, so for my points, I get x =11.045 and x=3.6215, we have x=0 and x=12 that we need to test as well because those are endpoints, you just have to find the other 2 critical points using the quadratic formula. Do you know what to do next with the candidates test ?

Answer 11

@snowflake0531 snowy

Answer 12

\[3t^2-44t+120=0\]
\[t=\frac{ 44 \pm \sqrt{(-44)^2-4\times 3\times 120} }{ 2\times3}\]
\[t=\frac{ 44 \pm \sqrt{1936-1440} }{ 6 }=\frac{ 44 \pm \sqrt{496} }{ 6 }=\frac{ 44 \pm 2\sqrt{124} }{ 6}\]
\[t=\frac{ 22 \pm \sqrt{124} }{ 3 }\]
\[H''(t)=t-\frac{ 22 }{ 3}\]
when \[t=\frac{ 22-\sqrt{124} }{ 3 }\]
\[t-\frac{ 22 }{ 3 }=-\frac{ \sqrt{124} }{ 3 }\]
\[H''(t)=t-\frac{ 22 }{ 3}\]
\[=-\frac{ 124 }{ 3 }<0\]
so H(t) is maximum at\[t=\frac{ 22-\sqrt{124} }{ 3 }\]
when \[t=\frac{ 22+\sqrt{124} }{3}=\frac{ 22 }{ 3 }+\frac{ \sqrt{124} }{ 3 },\]
\[H(t)=\frac{ \sqrt{124} }{ 3}>0, \]
H(t) is minimum at this point.
we need to find H(t) at t=0,12 and \[t=\frac{ 22+\sqrt{124} }{ 3 }\]
and find which one is minimum.

Answer 13

that looks like some hard math questions

Answer 14

oh I’m so dumb I should have just done that

Answer 15

basically what i did was get the decimal version, get the endpoints, then it was easy