TOPIC: anyone smurf at maths

"an enclosure is in the shape of a rectangle. one side of the enclosure is fenced in by a building. 100m of fencing is available to fence the other 3 sides"
(I)"if the width of the enclosure is x m, find in terms of x, the lenght of the enclosure"
(II)"show that the area of enclosure is given by A = 100x-2x^2"
(III)"Find the value of x that maximes the area of the enclosure"
(IV)"Find the maximum area of the enclosure"

area and volume / differentiation question please help.
I just don't know how to start it tbh

I would gladly help if I knew/understood any of the terminology

Davidov007 wrote: david tried to help me on my maths test and i got half the pass score

jam wrote: sorry but people that don't understand the formula of an area of a circle can't be helped by anyone

Davidov007 wrote:
what don't you understand

MichaeI wrote: DM me at Spooky Ace#3797 I'll try to help

Ask finaly he's a math boy

Assuming this is in a book can you send me a screenshot of it? It might be easier to understand

its fine david is 9000iq

Not a smurf in terminology sorry Michel

The shape is a rectangle but fencing is required only for 3 sides, 2 widths and 1 length. Question specifies that x m is a width, so we can give length any variable (y etc.) . Fencing is perimeter so it will be / 2x + y = 100 (100 being the avaible number of fencing) / . In terms of x means that you need to put y on its own => y = 100 - 2x


The area forumla is / Area = width * length ( A = xy ) / . You have an equation from (i), so just fill in the Area equation. A = x(100 - 2x) == 100x - 2x^2


You have the basic done from (ii). You need to use calculus to differentiate the area equation and find the maximum, not the minimum!, which can be found in your log/text book.

Studio Banter wrote: Not to be a dick but you did just waste your time because he mentioned that he got the help he needed.
Good job with the solving, though.