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

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

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

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

this video just saved my life. thank you!

saved my grade props man

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.

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

Saving me the night before test

I have found my new math wizard

Bro bless man

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 :)

Thank you so much you seriously saved me

Wow this is great! Thanks!