Optimization question

Question

Find two positive numbers satisfying the given requirements.

The sum of the first and twice the second is 200 and the product is a maximum

first number=?

second number=?

Answer 1

First number plus twice Second number = 200

F+2S = 200

F=200-2S

FS = 200-2S)S= 200S -2S^2

take the derivative and set equal to zero

200-4S = 0

4S = 200

S =200/4 = 50 = 2nd number

F=200-2S= 100 = first number

product = 5,000

try 2 other numbers, say 150 = 1st number and 2nd =25

1st + twice 2nd = 200

their product is 150x25= 3750

or try 90 and 55, product = 4950

or 110 and 45, 4950

either side of 100 and 50 gets a smaller product

190 and 5, 190+2(5) = 200, 190x5= 950

the further away from 100 and 50, the smaller the product

198 and 1, product is 198

199 1/2 and 1/4, product is 49.875

200 and 0, product = 0, 200+2(0) = 200

200 and 0 are two numbers that minimize the product while the first plus 2 times the 2nd sum to 200

also 0 and 100 minimize the product and 0 +2(100) = 200

every other pair of numbers' products will be between 0 and 5,000

Answer 2

Let the numbers be x and y

Then, if the sum of the first and twice the second number is 200, x + 2y = 200

We seek to maximize the product, xy

Since x + 2y = 200, x = 200 - 2y

Then, we seek to maximize

(200 - 2y)y =

200y - 2y^2 =

-2(y^2 - 100y) =

-2(y^2 - 100y + 50^2) + 2(50^2) =

-2(y - 50)^2 + 5000

This is maximized at y = 50 and has a maximum value of 5000

x = 200 - 2y = 200 - 2(50) = 100