Calc I Min/Max Problem?

You have 2 numbers, x and y

x^2 + y = 60

y = 60 - x^2

P = x * y

P = x * (60 - x^2)

P = 60x - x^3

Derive P with respect to x

dP/dx = 60 - 3x^2

dP/dx = 0

0 = 60 - 3x^2

3x^2 = 60

x^2 = 20

x = +/- 2 * sqrt(5)

x > 0, so x = 2 * sqrt(5)

x^2 + y = 60

20 + y = 60

y = 40

Let x and y be the two numbers

x^2 + y = 60

Product = xy is maximum

Product = xy

Express y in terms of x

Product = x (60-x^2)

Product = 60x -x^3

dP/dx = 60 -3x^2 = 0

3x^2 = 60

x^2 = 20

x = ± √20

d^2P/dx^2 = -6x

when x= √20, d^2P/dx^2 = -6 (√20) < 0 , so Product has a maximum at x= √20

The numbers are x = √20

y= 60 - x^2 = 60- 20 = 40

The numbers are √20 and 40

The maximum product is : (40)(√20) = 178.8854

Let the numbers be x and y

→x²+y=60

i.e. y=60−x²

Product P=xy

P=xy=x(60−x²) =60x−x³

dP/dx= 60−3x² =0 for maxima or minima

→x²=20and x=√20=2√5 Since numbers are positive

Hence y=(60−x²)=(60−20)=40

d²P/dx²= −6x negative for positive x Hence the Value of x is for maximum P

Maximum product P=xy=2√5×40=80√5

x > 0, y > 0

x² + y = 60

y = 60 - x²

find max P = xy

= x(60 - x²) = 60x - x³

P' = dP/dx = 60 - 3x² = 0

3x² = 60

x² = 20

x = ±√20, reject negative value

x = √20

y = 60 - x² = 60 - 20 = 40

y = 40

P = xy = 40√20

x = √20, y = 40

P" = - 6x

-6√20 < 0

Therefore P is a maximum.