# optimization problems

• Mar 24th 2007, 08:41 PM
juiicycouture
optimization problems
1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?
• Mar 24th 2007, 09:07 PM
Jhevon
Quote:

Originally Posted by juiicycouture
1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?

ALWAYS START OPTIMIZATION PROBLEMS BY DRAWING DIAGRAMS! see below, i drew some for you.

Quote:

1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?
2x + 2y = 16 = P ..........P is for perimeter

we want Area = A = xy to be a maximum

since 2x + 2y = 16
=> x = 8 - y

so A = (8 - y)y = 8y - y^2

for max, we want A' = 0
=> A' = 8 - 2y = 0
=> y = 4

But x = 8 - y
=> x = 4

Quote:

2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?
this is very similar to the last one

x + 2y = 40

want A = xy to be max

since x + 2y = 40
=> x = 40 - 2y

=> A = (40 - 2y)y = 40y - 2y^2

for max, A' = 0
=> A' = 40 - 4y = 0
=> y = 10

but x = 40 - 2y
=> x = 20

Quote:

3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?
A = xy = 32

we want Perimeter = P = 2x + 2y to be minimum

since xy=32
=> x = 32/y

=> 2(32/y) + 2y = P
=> P = 64/y + 2y = 64y^-1 + 2y

for min, P' = 0
=> P' = -64y^-2 + 2 = 0
=> -64 + 2y^2 = 0
=> 2y^2 = 64
=> y^2 = 32
=> y = +/- sqrt(32)
so y = sqrt(32) since we can't have negative area

but x = 32/y
=> x = 32/sqrt(32)
=> x = sqrt(32)
• Mar 24th 2007, 09:19 PM
Jhevon
Quote:

Originally Posted by juiicycouture
1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?

hold on, you're in pre-calculus?

that means you don't know how to find a derivative right?
• Mar 24th 2007, 09:19 PM
ecMathGeek
Quote:

Originally Posted by juiicycouture
1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?

juiicycouture, do you know what derivatives are and how to use them? I suspect that you're are in Precalculus, and so you probably haven't learned Calculus yet. If this is true, say so and I, or one of the other helpers on this site, will show you how to do these problems without derivatives.
• Mar 24th 2007, 09:20 PM
ecMathGeek
Quote:

Originally Posted by Jhevon
hold on, you're in pre-calculus?

that means you don't know how to find a derivative right?

lol
• Mar 24th 2007, 09:20 PM
Jhevon
Quote:

Originally Posted by ecMathGeek
juiicycouture, do you know what derivatives are and how to use them? I suspect that you're are in Precalculus, and so you probably haven't learned Calculus yet. If this is true, say so and I, or one of the other helpers on this site, will show you how to do these problems without derivatives.

yeah, sorry, i realized afterwards that this thread was in precalc.

we will need to use the vertex equation then
• Mar 24th 2007, 09:22 PM
Jhevon
ah well, juiicycouture went offline. you wanna do the precalc way anyway ecMathGeek? or should i? 'cause you seem bored
• Mar 24th 2007, 09:40 PM
Jhevon
Quote:

question 1
let's start from

so A = (8 - y)y = 8y - y^2

for max, we want y = -b/2a .........this is the formula for the vertex of the parabola
=> y = -8/2(-1) = -8/-2
=> y = 4

But x = 8 - y
=> x = 4

Quote:

question 2
let's begin from here
=> A = (40 - 2y)y = 40y - 2y^2

for max, y = -b/2a
=> y = -40/2(-2) = -40/-4
=> y = 10

but x = 40 - 2y
=> x = 20
• Mar 24th 2007, 10:28 PM
ecMathGeek
Quote:

Originally Posted by juiicycouture
1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?

3)
The deminsions we will use are x = length, and y = width
We are given A = 32 = xy
We need P = 2x + 2y to be a minimum.

First, let's solve 32 = xy for x (we could solve it for y, too... we'd get the same answer)
32 = xy
x = 32/y

Second, plug in x = 32/y into P = 2x + 2y
P = 2(32/y) + 2y
P = 64/y + 2y
P = 64y^(-1) + 2y ... since 1/y = y^(-1)

Now, we need to make this a quadratic function (which we can do if we multiply everything by y)
yP = 64 + 2y^2
2y^2 - Py + 64 = 0

For the minimum area, all we need is y = -b/(2a)
y = -(-P)/(2*2)
y = P/4

Since P = 2y + 2x
2x = P - 2y
x = P/2 - y ... Plug in y = P/4
x = P/2 - P/4
x = P/4

Therefore, x and y are one forth the total perimater.
• Mar 24th 2007, 10:35 PM
ecMathGeek
Quote:

Originally Posted by ecMathGeek
3)
The deminsions we will use are x = length, and y = width
We are given A = 32 = xy
We need P = 2x + 2y to be a minimum.

First, let's solve 32 = xy for x (we could solve it for y, too... we'd get the same answer)
32 = xy
x = 32/y

Second, plug in x = 32/y into P = 2x + 2y
P = 2(32/y) + 2y
P = 64/y + 2y
P = 64y^(-1) + 2y ... since 1/y = y^(-1)

Now, we need to make this a quadratic function (which we can do if we multiply everything by y)
yP = 64 + 2y^2
2y^2 - Py + 64 = 0

For the minimum area, all we need is y = -b/(2a)
y = -(-P)/(2*2)
y = P/4

Since P = 2y + 2x
2x = P - 2y
x = P/2 - y ... Plug in y = P/4
x = P/2 - P/4
x = P/4

Therefore, x and y are one forth the total perimater.

I forgot to finish up this problem in solving for x and y (thanks Jhevon)

Since y = P/4 and x = P/4, we can say y = x. We can also plug this relationship back into 32 = xy
x^2 = 32
x = Sqrt(32)
AND
y = Sqrt(32)