For the calculus solution, you don't actually need to take the derivative or use calculus. Since the function is a downwards parabola, the x that makes the function the highest will be -b/2a, or in this case, -100/-4, giving 25 for x. you can then solve for the area and width fairly easily.

keep it close to 1/2 is for me you can do this to other numbers as well for example the sum must be 20 so the lenght must be unaffected since the other side doesn't need fencing so 10 ×5

Loved the video. You can also solve it by algebra. Find a quadratic function for the area in terms of L or W. Then find the maximum value by finding the vertex.

Here is a quicker method. The area is given by the formula A=L(100-2L). This clearly a quadratic and a quadratic has a stationary point halfway between the roots. The roots are at 0 and 50 So the stationary point is halfway between at L=25. The area at the point is positive(1250). So the stationary point is a maximum.

At 3:18 or so he states that the area is larger by X squared. I have no idea how he came by that. You lose the small area X.L and you gain the area X(W-X). So the total area gained would be X(W-L)-X^2.

This fence problem was literally my first introduction to calculus from a friend of mine who had started learning it on his own when in 7th or 8th grade. — The 'build a hypothetical fence in the water' solution is new to me however.

I also considered a semi-circle which will give you the largest most optimized area of 1592. You didn't specify that it must be a rectangle in your original question, though that's the solution you proceeded with, a few others came up with the same answer I did. Start with a circle of perimeter 200, calculate your radius, then area then divide by 2.

You can also solve the problem geometrically without proving that a square is optimal for a complete rectangle.

Take any solution w*l. To increase l by e, you need to decrease w by 2e – which is similar to shaving a strip e wide off each vertical side, and placing them side-by-side on top of the rectangle – and decreasing l by e looks like shaving an e high strip off the top, splitting it in half at the midpoint, then adding those two strips to the two ends.

Moving the strips around preserves the total area, and changes the length and width appropriately, but the strips will, in general, either be too long or too short to fit the new area perfectly. If w<2l, then reducing l will add shorter strips to either side, meaning the area increases, while increasing l will give an overlap at one or both ends meaning the area decreases. Similarly, if w>2l, reducing l means the strips on either side will stick up above the top of the new rectangle (meaning area is reduced) while increasing l means the two strips won't cover the full width (so area increases).

Since area increases as you move toward w=2l from either direction, that must be the maximum.

Area = 100L – 2L^2, this is an upside down parabola function. Therefore the max area is at its vertex. L = -b/2a = -100/2(-2) = 25, hence W = 50. A= LW = 100L-2L^ = 25 * 50 = 100(25)-2(25^2) = 1250

This is how students solve it in intermediate algebra class

I used an old formula I discovered as a kid: (x-y)(x+y)=x^2-y^2

It's not just good for working out areas from a change length or width with a fixedperimeter It's a great technique for mental multiplication when already know the square of the number's averages. For example. 997 * 1003 x = 1000 & x^2 = 1,000,000 y = 3 & y^2 = 9 1,000,000 – 9 = 999,991

The correct answer is L=1 and W=98 because the majority of area in all other solutions isn't directly next to the river (being that it is one unit away at least).

I knew I had to optimize so I just located the vertex (f(x) was area and x was length) to get the best length. Then back substitution to get width. No calculus.

## 42 Replies to “Math Puzzler: Optimize The Fence”

I will assume that each of the hundred units is indivisible and straight, and make a half regular 200-gon

i solved it in a Another way.

For the calculus solution, you don't actually need to take the derivative or use calculus. Since the function is a downwards parabola, the x that makes the function the highest will be -b/2a, or in this case, -100/-4, giving 25 for x. you can then solve for the area and width fairly easily.

keep it close to 1/2 is for me you can do this to other numbers as well for example the sum must be 20 so the lenght must be unaffected since the other side doesn't need fencing so 10 ×5

Side lengths 33×34

a triangle with fence lengths of 50 perpendicular to each other would maximize the waterfront and the area simultaneously.

The largest area wouid be a semicircle fence .

πR = 100

–> R = 100/π

Area = πRR/2 = π.100.100/π.π.2

= 10000/2π

= 1591.5

hey presh. please make more

Loved the video. You can also solve it by algebra. Find a quadratic function for the area in terms of L or W. Then find the maximum value by finding the vertex.

It's 1591!!

Here is a quicker method.

The area is given by the formula A=L(100-2L).

This clearly a quadratic and a quadratic has a stationary point halfway between the roots. The roots are at 0 and 50

So the stationary point is halfway between at L=25.

The area at the point is positive(1250). So the stationary point is a maximum.

I am doing this in calculus

At 3:18 or so he states that the area is larger by X squared. I have no idea how he came by that. You lose the small area X.L and you gain the area X(W-X). So the total area gained would be X(W-L)-X^2.

2:00

Why do you do 100 – 4L

This fence problem was literally my first introduction to calculus from a friend of mine who had started learning it on his own when in 7th or 8th grade. — The 'build a hypothetical fence in the water' solution is new to me however.

I also considered a semi-circle which will give you the largest most optimized area of 1592. You didn't specify that it must be a rectangle in your original question, though that's the solution you proceeded with, a few others came up with the same answer I did. Start with a circle of perimeter 200, calculate your radius, then area then divide by 2.

I did (10+20+30+40)/4 to get 25 as the length

The best way would be to make a half circle wiht radiu of 31,84 units

I just did 100/2

Then 50/2

How will you justify that its a quadrilateral?? It might be any other shape

Do you realise, your example in the question was totally misleading???

you just need a small brain to solve this in 10 sec without any calculations

If you make it 3/4 of a circle you'll get a lot more area

I was having flashbacks of growing up on a cattle farm.

You can also solve the problem geometrically without proving that a square is optimal for a complete rectangle.

Take any solution w*l. To increase l by e, you need to decrease w by 2e – which is similar to shaving a strip e wide off each vertical side, and placing them side-by-side on top of the rectangle – and decreasing l by e looks like shaving an e high strip off the top, splitting it in half at the midpoint, then adding those two strips to the two ends.

Moving the strips around preserves the total area, and changes the length and width appropriately, but the strips will, in general, either be too long or too short to fit the new area perfectly. If w<2l, then reducing l will add shorter strips to either side, meaning the area increases, while increasing l will give an overlap at one or both ends meaning the area decreases. Similarly, if w>2l, reducing l means the strips on either side will stick up above the top of the new rectangle (meaning area is reduced) while increasing l means the two strips won't cover the full width (so area increases).

Since area increases as you move toward w=2l from either direction, that must be the maximum.

This is just differntiation!

Are you Indian??????

Oh, a square box

wouldbe the largest area if you were making all four sides, but because it's only three, the answer changes.Area = 100L – 2L^2, this is an upside down parabola function. Therefore the max area is at its vertex. L = -b/2a = -100/2(-2) = 25, hence W = 50. A= LW = 100L-2L^ = 25 * 50 = 100(25)-2(25^2) = 1250

This is how students solve it in intermediate algebra class

I dont understand how u wouldnt cover more area of the river if you made the walls 1 (or even .1) and the width 98 (or 99.8)

semi-circle if there wasnt the "wire-fencing draw it as a straight line basically rectangle form" requirement

I missed the "3 sides of a box" constraint – a semi-circular fence encloses ~1591 square units, the rectangular area maxes out at 1250 units squared.

Build half a 200-gon with the 100 edges of 1 unit each. The area is 25 * cot (π/200).

25+25+50=100, 25*50=1250

I used a graphing calculator

I used an old formula I discovered as a kid:

(x-y)(x+y)=x^2-y^2

It's not just good for working out areas from a change length or width with a fixedperimeter It's a great technique for mental multiplication when already know the square of the number's averages. For example.

997 * 1003

x = 1000 & x^2 = 1,000,000

y = 3 & y^2 = 9

1,000,000 – 9 = 999,991

You should be on Trump's wall building committee.

The correct answer is L=1 and W=98 because the majority of area in all other solutions isn't directly next to the river (being that it is one unit away at least).

Wouldn't a half circle be better? Or even best?

What shape maximizes the area that can be enclosed with 100 units of fencing?

The second way is how i always want to solve the problem in calculus but my teacher doesnt let me 😭

Geerkens' Law: Sum of horizontal fencing = sum of vertical fencing at maximum area. So maximum area = 50 * 50/2 = 1250.

I knew I had to optimize so I just located the vertex (f(x) was area and x was length) to get the best length. Then back substitution to get width. No calculus.