Assuming the two numbers are x and y so that x+y = 40 Where xy is maximum Solve y in terms of x y = 40-x Then since f(x,y) = xy is maximum substitute y = 40-x f(x,y) = x(40-x) Get the derivative both sides in terms of x f'(x,y) = 40-2x Equate it to 0 to solve x 0 = 40-2x 2x = 40 x = 20
For x+y = 40 Substitute x = 20 20+y = 40 y = 40-20 y = 20
Therefore The two numbers are 20 and 20 whose their product is in maximum
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Assuming the two numbers are x and y so thatx+y = 40
Where xy is maximum
Solve y in terms of x
y = 40-x
Then since
f(x,y) = xy is maximum substitute y = 40-x
f(x,y) = x(40-x)
Get the derivative both sides in terms of x
f'(x,y) = 40-2x
Equate it to 0 to solve x
0 = 40-2x
2x = 40
x = 20
For x+y = 40
Substitute x = 20
20+y = 40
y = 40-20
y = 20
Therefore
The two numbers are 20 and 20 whose their product is in maximum