Question

Please help! Will give medal. A rectangular corral is to be built with a required area, k. If an existing fence is to be used as one of the sides, determine the dimension of the corral which would cost the least.

Answer 1

l = length
w = width
k = area

Answer 2

Any more information? @danielle02

Answer 3

@iGreen That's all I got. I have to solve this using calculus but I don't know how to start.

Answer 4

we are given that
x*y = k

and lets say it costs c dollars per foot of fence

total cost = c*2x + c*y

Answer 5

@perl So how do I find the dimensions?

Answer 6

you can factor out c :
Cost = c*(2x + y)

Answer 7

cost (x) = c * ( 2x + y)
we are given that
x*y = k, so y = k / x
substitute

cost(x) = c * ( 2x + k/x )

Answer 8

we can minimize that function , since it is in one unknown

Answer 9

find the derivative of that

Answer 10

how did it become f'(x)=c[2-(k/x^2)] ?

Answer 11

$$
\Large f(x) = c * ( 2x + \frac{k}{x} ) \\
\Large f(x) = c * ( 2x + k \cdot x^{-1} ) \\
\Large\color{red}{\Large f~'(x) = c * ( 2 \cdot 1 + k \cdot (-1) x^{-2} ) }\\

\Large f ~' (x) = c * ( 2 - \frac{k}{x^2})\\

\Large f ' (x) = 0 \iff c * ( 2 - \frac{k}{x^2}) = 0 \\

\text {divide both sides by c}\\

\Large 2 - \frac{k}{x^2}
=0\\
\Large 2 = \frac{k}{x^2}\\

\Large 2x^2 = k \\
\Large x^2 = \frac{k}{2} \\
\Large x = \pm \sqrt{\frac{k}{2}} \\
\text {since length is positive}\\
\Large x= \sqrt{\frac{k}{2}}
$$

Answer 12

So that would be the dimension? Or do I still need to work on another equation?

Answer 13

in the drawing how would you define length?

Answer 14

we found the dimension that minimizes cost is x = sqrt(k/2)
is this width?

Answer 15

to me , length and width are ambiguous

Answer 16

@perl Thank you for helping me understand the problem. I appreciate all your effort. Thanks again!

Answer 17

There was a hint that the width= twice the length

Answer 18

yes , do you see why?

Answer 19

$$
\text {we found that}\\
\Large x= \sqrt{\frac{k}{2}} = \frac{\sqrt{k}}{\sqrt{2}} = \frac{\sqrt{k}} {\sqrt{2}}\cdot\color{red}{\frac{\sqrt{2}}{\sqrt{2}}}= \frac{\sqrt{2}\cdot \sqrt{k}}{2}
\\ \Large \text {but}
\\ \Large x \cdot y = k
\\ \Large (\frac{\sqrt{2}\cdot \sqrt{k}}{2}) \cdot y =k
\\ \Large \text{ so }
\\ \Large y = k \cdot (\frac{2}{\sqrt{2}\cdot \sqrt{k}}) = \frac{2k}{\sqrt{2k}}\cdot \color{red}{\frac{\sqrt{2k}}{\sqrt{2k}}}= \frac{2k\cdot \sqrt{2k}}{2k}\\
\\ \Large = \sqrt{ 2k} = \sqrt{2}\cdot \sqrt{k}


$$

Answer 20

now you see x is half of y (or if we switched the roles of x and y, then y is half of x)