ithe fence problem can be solved by polynomials l*w=surface l+2w = 1000 l=1000-2w f(w)= surface = w(1000-2w) f(w) = -2w+1000w The max value of f(w) and w is in the point A = {-b/2a;-∆/4a) the max value of f(w) = 1000²/8 = 125000 m² the value of w at which we get this f(w) is 1000/4=250 L=500 W=250
That's correct. It is basically the same as finding the vertex of a parabola. I think I mentioned in the video that it can be solved algebraically. The point to note is that many max-min problems require calculus, and that is one of the types of problems that motivated the development of calculus historically.
ithe fence problem can be solved by polynomials
l*w=surface
l+2w = 1000
l=1000-2w
f(w)= surface = w(1000-2w)
f(w) = -2w+1000w
The max value of f(w) and w is in the point A = {-b/2a;-∆/4a)
the max value of f(w) = 1000²/8 = 125000 m²
the value of w at which we get this f(w) is 1000/4=250
L=500
W=250
That's correct. It is basically the same as finding the vertex of a parabola. I think I mentioned in the video that it can be solved algebraically. The point to note is that many max-min problems require calculus, and that is one of the types of problems that motivated the development of calculus historically.