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## Homework Statement

Among all the pairs of numbers with a sum of 236, find the product whose pair is maximum.

## Homework Equations

## The Attempt at a Solution

I got the correct solution to this, but I feel it was a weak way to do it and I'd like to know how to do it more efficiently. For example this way wouldn't work if it were more than two numbers.

Work:

(i) Let x= the first number and Let y= the second number

(x+y) = 236 and xy = maximum product

(ii) Then I used each number as somewhat of a side of a rectangle in that:

1/2Perimeter of Rectangle = x+ y = 236

(iii) I then redefined the rectangle in terms of y. That is,

Width = y and Length = 236-y

(iii) Put into an equation for area, as Area of Rectangle= lw which I held as synonymous to the given question in that:

Maximum Area of Rectangle = (l)(w) ⇒ product of two numbers with a maximum result (a)(b)

1/2Perimeter of Rectangle = l + w ⇒ two numbers whose sum is 236

(iv) Finding a maximum area of a rectangle ⇒ finding a maximum product of two numbers

a(y) = y(236-y)

a(y) = -y^2+236y

a(y) = -(y^2+236y+13924)+13924

a(y) = -(y-18)^2+13924

Maximum value at vertex = (118,13924)

(v) Maximum area of rectangle with sides that add up to 236 would be sides lengths of 118 and 118

So the two numbers whose sum is 236 and sum is maximum would be

x=118 and y=118.

What is a more efficient way to do this problem that doesn't restrict me to two numbers?

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