Question

math problem

Answer 1

@AZ

Answer 2

this is a very interesting question hmmm

Answer 3

And we can't use calculus either so let's dumb it down



what is the perimeter?

add up all the letters that you see

Answer 4

wouldnt it just be 1 x on either side bc it's 4 walls? and not 6

Answer 5

the perimeter = 234

and we also know that the area is length * width

and we have two rectangles here so 2 * length *width

the length is x
and the width is y

so area = 2 * x * y

and we want to get the maximum area

but we can proceed once you determine what the formula for perimeter is

Answer 6

let x be the length and y be the width.
234=2x+3y
3y=234-2x
\[y=\frac{234-2x}{3}\]
\[area A=x*y=x(\frac{234-2x}{3})\]
\[A=\frac{1}{3}(234 x-2x^2)\]
\[\frac{dA}{dx}=78-\frac{4x}{3}\]
\[\frac{dA}{dx}=0~gives~4x=234\]
x=117/2
\[\frac{d^2A}{dx^2}=-\frac{4}{3}\]
\[\frac{d^2A}{dx^2} <0~at~x=117/2\]
so A is maximum if x=117/2
3y=234-2 *117/2=234-117=117\]
y=117/3=39
so with the given condition
length=117/2=58.5 ft
width =39 ft

Answer 7

thanks

Answer 8

yw

Answer 9

Hmm, yeah that is correct. I could have just made the entire length as 'x' instead of breaking it up like I did in my drawing