Using Quadratics to Quickly Find the Maximum Area of a 3 Sided Fence Given a Certain Perimeter

[@Faytee851]

year old video helping me more than my math teacher, props

[@Nabifroese]

An oldy but a goody? Lol! Glad it helped!

[@thezj_5859]

Thank you so much! I have an exam tomorrow and had no idea how to figure out these questions. We haven't learned calculus so I don;t know how to use "derivatives", so thank you for showing how to do it this way. Sub from Australia

[@Nabifroese]

You're very welcome!

[@lydia4862]

this video just saved my life. thank you!

[@Nabifroese]

I'm so glad!

[@aleksandaruzunov5726]

great video, thanks for bringing things back ...things I already forgot. At the end of this problem, instead of the point 20,40 it should be (20,800) because the second coordinate is actually the Area, not the lenth

[@Nabifroese]

Thanks for your point, no pun intended! In this question they just asked for the best possible dimensions which were indeed 20 and 40... however if we were to graph this the maximum y value on the parabola would be 800 as you pointed out. Thanks!

[@notasinglesoul1179]

saved my grade props man

[@Nabifroese]

Nice!

[@MennaHany-e5f]

Let’s say the length of the rectangular lot is x and the width is y. Since one side of the lot is bounded by a river, only three sides need to be enclosed by fencing. We can write an equation for the perimeter: 2x + y = 80. Solving for y, we get y = 80 - 2x. The area of the rectangle is given by A = xy. Substituting the expression for y from above, we get A = x(80 - 2x) = 80x - 2x^2. To maximize the area, we need to find the value of x that makes A as large as possible. Taking the derivative of A with respect to x, we get dA/dx = 80 - 4x. Setting this equal to zero and solving for x, we find that x = 20. Substituting this value back into our expression for y, we find that y = 40. So, the dimensions of the largest lot possible are 20 meters by 40 meters, and the maximum area that can be enclosed by the fence is A = xy = 20 * 40 = 800 square meters.

[@Nabifroese]

I like your explanation! The only issue is that most high school students have not yet learned calculus to solve this problem. But I agree, the derivative is a fun tool to use in this situation!

[@nicojones796]

Saving me the night before test

[@Nabifroese]

☺️

[@beaurichards6146]

perfect vid thanks heaps you've just added +1 mark to my test

[@Nabifroese]

Nice!

[@stickbro4118]

I have found my new math wizard

[@Nabifroese]

😎👍

[@Asianpotato77]

Bro bless man

[@Nabifroese]

Glad to help!

[@user-nemyong]

Thank you so much you seriously saved me

[@Nabifroese]

Awesome!

[@djlucius7641]

I know this video is 2 years ago but if I can get an answer on why the Vertex's X Value is the ACTUAL width of the rectangle instead of 40 I would appreciate it a lot :)

[@Nabifroese]

I'm sorry I am just on holidays right now but will try to have a look at your question in the near future my apologies!

[@Nabifroese]

The reason the width is not 40 is because using quadratics we found that there are two x-intercepts or zeros. One is x = 0 and the other one is x = 40. To find the maximum wdth we need to find the vertex of the parabola and this vertex would happen in between 0 and 40. The maximum width possible would be 20 because that number is in between the two x intercepts. Sorry I am just quickly talking into my phone helping with this question rather than sitting down and typing it out. It's the best I can do from the situation I am in right now lol. Hope this helps a little bit.

[@santiagoortega5625]

Wow this is great! Thanks!