can u help me with his calculus problem plz?

Question

A landscape architect is creating a rectangular rose graden to be located in a local park. the rose garden is to have an area of 60m^2 and be surrounded by a lawn. the surrounding lawn is to 10 m wide on the nothe and south sides of the garden and 3 on the east and west side. Find the dimensions of the rose garden if the total area of the garden and lawn together is to be a minimum

Answer 1

let the dimension of the rose garden be x*y
given xy=60
so x=60/y
area of the lawn androse garden together=(x+10)(y+3)
substitutingthe value of x intermsof y=(60/y +10)(y+3)
=60+180/y+10y+30
dA/dy=-180/y^2+10
setting this to 0
-180+10y^2=0
y^2=18
y=rt18=3rt2
substituting x=60/3rt2=10rt2
so the dimensions of the rose garden
length=10rt2m
and breadth=3 rt2 m

Answer 2

Let

x = length of rose garden
y = width of rose garden
A = area of garden and lawn

GIven

xy = 60 m²

Surrounding lawn is 10m wide on the north and south, and 3m wide on the east and west.

Find

x and y to minimize area of garden and lawn.

xy = 60
y = 60/x

A = (x + 2*10)(y + 2*3) = (x + 20)(y + 6)
A = xy + 6x + 20y + 1200 = 60 + 6x + 20(60/x) + 1200
A = 6x + 1200/x + 1260

dA/dx = 6 - 1200/x²
6 = 1200/x²
6x² = 1200
x² = 200
x = 10√2

y = 60/x = 60/(10√2) = 6/√2 = 6√2/2 = 3√2

So the dimensions of the rose garden are
10√2m by 3√2m

Answer 3

dimensions of the rose garden
length = 10* (2)^.5
breadth = 6/(2)^.5
let x be breadth of rose garden
y be length of rose garden
x*y=60 --(1)
A = (x+6)(y+20) ---(2)
now substitute x in eq. 2 from eq 1
differentiate A , put it equal to 0
find y