Can u help me with his calculus problem plz?

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

Can u help me with his calculus problem plz?
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
Reply: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
Reply: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