Question

Can someone please look at the attachment

Answer 1

what attachment?

Answer 2

call one side \(x\). at $3 a foot it will cost \(6x\)
and the other side \(y\) and at $5 a foot it will cost \(10y\)
for a total cost of \(C=6x+10y\)

then since \(xy=3000\) you have \(y=\frac{3000}{x}\)
now the cost in terms of \(x\) will be
\[C(x)=6x+10\times \frac{3000}{x}\] minimize that in the usual calculus method

Answer 3

I am confused by what you mean minimize it

Answer 4

@satellite73 what do I plug in for the x value

Answer 5

can you please help me I am stuck what does the minimize part mean

Answer 6

your job is to find the minimum value of the function
your posts are almost always calculus so i am assuming that this is for a calc class
am i right?

Answer 7

yes and my teacher is not very good so I am confused over half the time so I don't understand the minimize part what you are asking me to do

Answer 8

actually it must be because you cannot do this just using algebra
your function is
\[C(x)=6x+\frac{30000}{x}\] and you want the minimum value
take the derivative, find the critical point (where the derivative is zero) and that will give your minimum

Answer 9

i can walk you through it if you like, it will not take too much

Answer 10

ok could you please because at this point I am totally lost sorry:(

Answer 11

have you seen a calc problem before where you were asked to find the maximum or minimum of a function?

Answer 12

like find the minimum of \[f(x)=x^2+2x-4\] by writing
\[f'(x)=2x+2\] then setting
\[2x+2=0\] to get \(x=-1\)

does this look familiar?

Answer 13

no we only had to find max and min on a graph

Answer 14

we are going to do exactly the same thing here
the function is
\[C(x)=6x+\frac{30000}{x}\] the hard part is finding the function, the rest is more or less routine
the derivative is
\[C'(x)=6-\frac{30000}{x^2}\] set that equal to zero and solve to find the critical points

Answer 15

you get
\[6-\frac{30000}{x^2}=0\]
\[\frac{30000}{x^2}=6\]
\[30000=6x^2\]
\[x^2=5000\]
\[x=\sqrt{5000}\]

Answer 16

so then final answer is 70.71

Answer 17

or if you prefer \(x=50\sqrt{2}\) or your decimal
don't forget this is \(x\) the length of the $3 side

Answer 18

ok

Answer 19

this is a typical calc max/min problem
you find the max or min by taking the derivative, finding the critical points (where the derivative is zero or undefined) and then checking to see if it is a max or a min

Answer 20

gotcha so just to make sure that I am following correctly the square root 5000 is the optimal dimmensions for the fence correct?

Answer 21

@satellite73 is this correct?

Answer 22

hold on

Answer 23

when you make your function you have to say what your variables represent

Answer 24

ok

Answer 25

i used \(x\) to represent the side of the fence that cost $3

Answer 26

ok so then do I take that70.71 and put it in the x place for c(x)=6x+(30000/x)

Answer 27

\(C(x)\) represents the cost, but you were asked for the dimensions
if you compute \(C(70.71)\) that will give the cost, but the dimensions are
\[x=50\sqrt{2}=70.71\] and
\[y=\frac{3000}{50\sqrt{2}}\] whatever that is

Answer 28

@satellite73 so the optimal dimensions are the x and y do I need to add them together