Question

Can someone answer this while showing the working out for this question?

Answer 1

The minimum cost is $`420` when AB`=15` m.

Explanation:

Let `AB=DC=x` meter.Then, `AD=BC=525/x` meter,
and the cost is
`f(x)=11x + 3(x+2*525/x)`
`=14x+3150/x` dollars. Needless to say, the domain is `x>0`.

Differenciate `f(x)`. Note that `d/dx(1/x)=d/dx(x^-1)=-x^-2=-1/x^2`.

`f'(x)=14-3150/x^2`

Now solve `f'(x)=0` to find the extrema.
`14-3150/x^2=0`
`14x^2-3150=0`
`x^2-225=0`
`(x+15)(x-15)=0`
`x=-15,15`

`x` is the length of `AB` and must be positive. So `x=15`.

And, when `0< x< 15`, `f(x)` is decreasing because `f'(x) <0`. When `15< x`, `f(x)` is increasing (`f'(x)> 0`).

Therefore, `f(15)` is the local (and global) minima and its value is `f(15)=14*15+3150/15=420`.

The minimum cost is `420` dollars.

graph{14x+3150/x [-3.15, 30, 400,500]}