Question

Medals shall be given to the best response :)
Please help, I have difficulty solving such questions.

A rectangular garden, 21m^2 in area, will be fenced to keep out rabbits and skunks. Find the dimensions that will require the least amount of fencing if a barn already protects one side of the garden.

Answer 1

First construct an equation(s) that model the problem.
You can draw a rectangle.
Let one of dimensions be x and one y (doesn't matter which one)
The perimeter is P = x + y + x (given one side, y, is the barn that already encloses the garden)
\[P = 2x+y\]
We also know the area is 21 sqm.
The area of the rectangle is A = xy
\[A = xy = 21, y=21/x\]
Now we can substitue y = 21/x into the perimeter formula.
\[P = 2x + y = 2x+\frac{ 21 }{ x }\]
All we need to do is find the first derivative and solve for zero as this tells us possible maxima or minima.
\[\frac{ dP }{ dx } = 2 - \frac{ 21 }{ x^2 } = \frac{ 2x^2 - 21 }{ x^2 } = 0\]
So we solve \[2x^2 - 21 = 0 -> 2x^2 = 21 -> x^2 = 21/2 -> x = \sqrt{\frac{ 21 }{ 2 }}\] taking x to be positive as it is a length.
Now you can confirm second differentiation that it is a minimum or check with first derivative around the point on both sides of the turning point.
Then substitute into any formula involving x to find y.

Answer 2

Thanks, but we are using methods in the advanced functions course. We can't use derivatives.

Answer 3

Are you allowed to sketch? If you do, you should see that adding the graphs of y=2x and
y =21/x together (which add to give P, the perimeter), the minimum occurs when they intersect for positive x.
So solve for 2x = 21/x and you will find such x.

Answer 4

Okay let me try it out

Answer 5

Could you explain the last step to me again. I don't understand it.

Answer 6

I end up with 2y^2 + 21/y = 0
When I solve that, we will have to take the square root of a negative. Which is impossible.

Answer 7

No no this is what I mean.
Sketch y =2x and y =21/x (very roughly) on the same axes and y = 2x +21/x
The intersection of y=2x and y=21/x is where y =2x +21x will be a minimum.
Compare:
http://www.wolframalpha.com/input/?i=2x+%3D+21%2Fx
http://www.wolframalpha.com/input/?i=2x+%2B+21%2Fx
So you solve NOT P = 0, but 2x = 21/x

Answer 8

Okay. The perimeter = 2x + y
Area = y = 21/x
I know where the y=21/x graph comes from, where do you get y = 2x?

Answer 9

From P =2x +21/x. to draw it you can add ordinates from y = 21x to y = 21/x.
It happens so when you add 2x to 21/x where they intersect, you get a minimum.

Answer 10

*y=2x to y =21/x

Answer 11

All i need to know is how you determined that y = 2x, that is the only part bothering me.

Answer 12

oh well perimeter is 2x + y right?
you also found y = 21/x...substitute into perimeter:
P = 2x + y = 2x + 21/x
So there's the 2x. All I did was draw 2x and 21/x on the same axes, then added points together to give 2x + 21/x.

Answer 13

Oh, so you treated 2x and 21/x as two seperate graphs?

Answer 14

yes. You don't need to but when you draw them separately but on the same set of axes, and then add points together, you find easily that the minimum occurs when y=2x happens to intersect with y =21/x.

Answer 15

Okay. Thank you very much :)

Answer 16

This might make more sense with all three graphs
http://www.wolframalpha.com/input/?i=graph%282x%2C+21%2Fx%2C+2x+%2B+21%2Fx%2C+x%3D-10..10%2C+y%3D+-20..20%29

Answer 17

Thanks.