Calc I Min/Max Problem?

Question

Can someone explain to me how to set this problem up? And what derivative am I looking for?

Find two positive numbers that satisfy the given requirements.
The sum of the first number squared and the second number is 60 and the product is a maximum.

Captain Matticus, LandPiratesInc · Accepted Answer

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

cidyah · Answer

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

Raj K · Answer

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

? · Answer

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.

Calc I Min/Max Problem?

4 個解答