Question

Calculus problem. Will award medal

Answer 1

where is the problem ?

Answer 2

http://gyazo.com/72577fe45056e21322b7b7a807bcc673

Answer 3

There ^

Answer 4

if we call with x, and y the sides of your fence, then we have:
area = x*y

Answer 5

this is a Lagrange multiplier or it can be made into a differential equation in 1 variable, depending on what level you are at.

Answer 6

so the perimeter of your fence, is:
perimeter=2(x+y)
now I substitute y with this value:
\[y = \frac{A}{x}\]
so we get:
\[perimeter = 2\left( {x + \frac{A}{x}} \right)\]

Answer 7

please note that the perimeter is a function of x only

Answer 8

so we have to minimize this function:
\[\Large f\left( x \right) = 2\left( {x + \frac{A}{x}} \right)\]

Answer 9

where A=180,000 square meters

Answer 10

what is the first derivative of f(x)?

Answer 11

please, do you know how to compute that first derivative?

Answer 12

I need to write this down first..yes I do

Answer 13

would I do the chain rule?

Answer 14

please you have to use the quotient rule, since I can rewrite that function as below:
\[\Large f\left( x \right) = 2\left( {x + \frac{A}{x}} \right) = 2\frac{{{x^2} + A}}{x}\]

Answer 15

ok

Answer 16

here is the first step:
\[\Large f'\left( x \right) = 2\frac{{2x \cdot x - \left( {{x^2} + A} \right) \cdot 1}}{{{x^2}}} = ...?\]

Answer 17

please wait a moment

Answer 18

so we can write:
\[\Large \begin{gathered}
f'\left( x \right) = 2\frac{{2x \cdot x - \left( {{x^2} + A} \right) \cdot 1}}{{{x^2}}} = 2\frac{{2{x^2} - {x^2} - A}}{{{x^2}}} = \hfill \\
= 2\frac{{{x^2} - A}}{{{x^2}}} \hfill \\
\end{gathered} \]

Answer 19

ok

Answer 20

now we have to apply this condition:
\[\Large f'\left( x \right) = 0\quad \Rightarrow 2\frac{{{x^2} - A}}{{{x^2}}} = 0\quad \Rightarrow {x^2} - A = 0\]

Answer 21

what is the acceptable value for x?

Answer 22

please note that we can get 2 possible values, nevertheless only one value is acceptable

Answer 23

A?

Answer 24

are you sure?

Answer 25

Nope

Answer 26

we have to solve this quadratic equation:
\[\Large {x^2} - A = 0\]

Answer 27

oh okay

Answer 28

which can be rewritten as below:
\[\Large \left( {x - \sqrt A } \right)\left( {x + \sqrt A } \right) = 0\]

Answer 29

now we have to apply the product cancellation law, so we gaet:
\[\Large \begin{gathered}
x - \sqrt A = 0 \hfill \\
x + \sqrt A = 0 \hfill \\
\end{gathered} \]
please solve both equation for x, what do you get?

Answer 30

\[x=\sqrt{A}\]

Answer 31

\[x=-\sqrt{A}\]

Answer 32

ok! and what is the acceptable solution?

Answer 33

please keep in mind that x is a length, so x has to be positive

Answer 34

X*Y=A?

Answer 35

we have
\[\Large x = y = \sqrt A \]
namely our fence is a square, and its perimeter, is:
\[\Large perimeter = 2\left( {\sqrt A + \sqrt A } \right) = 4\sqrt A \]

Answer 36

ok

Answer 37

we can show that perimeter is really the minimum perimeter

Answer 38

finally, I think that, we have to subtract one side, as requested from your problem ( no fence along the river), so the requested perimeter, is:
\[\Large perimeter = 4\sqrt A - \sqrt A = 3\sqrt A \]