Question

I have 2 homework questions that have really been giving me trouble.. if ANYONE can help with either I'd be so extremely grateful.

1) Find the global extrema of
f(x)= (1/3)x^3 - (1/2)x^2 + 6 on the interval [-3,4]


2) A farmer encloses a rectangular area so that area=200ft^2. The farmer will enclose three sides as there is already a fence on the fourth side. What dimensions should the rectangle be to use the least amount of fence? How much fence will be used?

Answer 1

Are you familiar with derivatives? You're going to need them for these two

Answer 2

Doulikepiecauseidont: yes and no.....this is my first calculus class and the professor is horrible. I know how to do the derivatives but in these problems am unsure of where to go

Answer 3

first of all you need to study derivatives because you don't expect us to answer your question because you won't understand a word we will say. It's easy once you learn it. Believe me. I've been there. and Done that.

Answer 4

Twinyang, if you would have read my previous comment you would see that I am familiar with derivatives. However, the process of solving the (above) question is new to me (I.e. I need some direction). Further, seeing as this is a public mathematics "help forum" I came for just that...help. If I understood all the aforementioned words^ I wouldn't be here in the first place.
Thank you for the helpful advice.

Answer 5

Notice that the given funciton is a polynomial, So the max/mins can occur only at two places :
1) when f'(x) = 0
2) at the given boundary : -3, 4

Answer 6

find the derivative of f(x), set it equal to 0 and solve x

Answer 7

\(\large f(x)= (1/3)x^3 - (1/2)x^2 + 6 \)

\(\large f'(x)=? \)

Answer 8

f'(x) = x^2 - x ?

Answer 9

yes ! set it equal to 0 and solve x

Answer 10

So, I did that and I factored out an x, --> looking something like x(x-1) = 0
and I got x= 1

But I'm not sure if that is correct and/or where to go from there

Answer 11

x(x-1) = 0

gives two solutions :
x = 0, 1

right ?

Answer 12

^ i.e where the interval comes in to play.... yes, two solutions! I'm sorry

Answer 13

So the max/mins can only occur at below `x` values :

0, 1, -3, 4

Answer 14

evaluate the given function at these x values,
then pick the MAX value for global max, and pick the MIN value for global min.

Answer 15

Do I plug those in to the original f(x)? or f'(x)?

Answer 16

\(\large f(x)= (1/3)x^3 - (1/2)x^2 + 6 \)

\(\large f(0) = ? \)
\(\large f(1) = ? \)
\(\large f(-3) = ? \)
\(\large f(4) = ? \)

Answer 17

f(0) = 6
f(1) = ~5.83
f(-3) = -7.5
f(4) = 19.33

So if these are correct then my global max would be at f(4) and the min at f(-3) ?

Answer 18

Exactly !

Answer 19

YES! That is so exciting. Thank you so, so much. I've really been struggling with the end of this assignment!

Answer 20

glad to see you figured it out :) yw !

Answer 21

Oh there is a second problem... lets see..

Answer 22

This is the problem: