A rectangular field is to be enclosed by 800 m of fencing. What dimensions will give the maximum area?
let A be one side B the other of the rectangular field. than we have that 2(a+b)=800 and we want to find the max for a*b
have you learnt differentiation?
Er, I don't think so. This is in the Quadratic functions unit
oh ok, so from the 2(a+b)=800 a+b=400 a=400-b now put this back to the area a*b=b(400-b)
is this clear until this, sorry that I am a bit slow, doing something else too :-)
why does a*b =b(400-b)
because a=400-b so we can swap the a for 400-b
Okay. So what do I do next?
? Where did the post go?
now we want to find the max for b(400-b)=400b-b^2
I deleted it because it was wrong :-)
do you know how to find the maximum of a function like this?
um. no. and I thought a*b = 400b-b^2
yeah thats right and we need to find for which value(s) of b will it have a max
okay. I'm confused. Go on though :p
Wait, isn't b(400-b)=400b-b the same as 400b-b^2=400b-b^2 ??/
ok you have to rewrite this expression to make a perfect square 400b-b^2= - (b^2-400b)= -((b-200)^2 -40000)
first step I took out the - so b^2 is positive
2nd step is to make a perfect square b^2-400b think about expanding the brackets for this (x+y)^2=x^2+2xy+y^2 so it will be (b+200)^2 as if you expand the bracket you get b^2+400b+40000 now that is 20000 more than our expression so we need to subtract that so b^2-400b=(b+200)^2 -40000
there is a typo, not 20000 but 40000
and now we are nearly done
because we want to find the maximum for this: -((b-200)^2 -40000)= =40000- (b-200)^2
do you see the solution?
OKAY, I had to write that all out for it to make sense, but in the end it did, :$. Thanks!
That is great!
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