Maxima Problems: “Find two possible real numbers whose product is a maximum, and the sum is 110 ?”
Answer
I’ll show you a couple of ways.
Let the two numbers be p and q respectively. We may further assume that both are positive. We are given that p + q = 110.
We want to determine reals p & q such that N = pq is maximum.
First. The “quadratic function” way.
N = pq
= p ( 110 – p ) from p + q = 100 we get q = 110 – p.
= 110p – p^2.
So N is a quadratic function in ‘p’. We may call it N(p).
We ‘complete the square’…’some manipulation’..
N (p) = 110p – p^2
= – [ p^2 – 110p ]
= – [ p^2 – 55p – 55p ]
= – [ p^2 – 55p -55p + 55^2 – 55^2 ]
= – [ p^2 – 55p -55p + 55^2 ] + 55^2
= – [ p ( p – 55 ) – 55 ( p – 55 ) ] + 55^2
= – [ ( p – 55 ) ( p – 55 ) ] + 55^2
= 55^2 – ( p – 55 )^2
= 3025 – ( p – 55 ) ^2 <= 3025 with equality if and only if
( p – 55 ) ^2 = 0 or p = 55.
That is N (p) < = 3025. Since we want to “maximize the product”, so N ( maximum ) = 3025 and this happens only when p = 55 as shown above. This q = 110 – 55 = 55. Hence the “possible reals are” : ( p, q) = ( 55, 55 ).
Note : “<=” above means “less than or equal to”. Similarly, “>=” will mean “bigger than or equal to”.
Method 2. ( Calculus )
We can “cut short” the method above via calculus using the idea that a “function” only attains its maximum of minimum when we set its derivative or dy/dx = 0. In our case,
N(p) 110p – p^2.
N(p) is the “function” while the “derivative” or “gradient function” or “dy/dx” is “symbolized” by N ‘(p).
Indeed,
N ‘ ( p ) = 110 – 2p. we set N'(p) = 0 to determine the ‘value of p for which the maximum for N(p) will occur ‘.
N'(p) = 0 implies that 110 = 2p and p =55 and we proceed as before to determine that (p,q) = ( 55,55) or p = q = 55.
Method 3. ( The AM – GM Inequality )
AM – GM : Arithmetic Geometric Mean Inequality.
Arithmetic mean of p,q, : ( p + q ) / 2
Geometric mean of p,q : sqrt (pq)
The AM – GM inequality theorem says that for any two ‘positive’ reals, p and q, the following holds with equality if and only if p = q. :
[ p + q ] / 2 >= sqrt ( pq )
Now since p + q = 110 , we get,
55 = 110/2 >= sqrt ( pq) or pq <= 55^2 = 3025. That is,
pq <= 3025. Now for pq ( max ) = 3025 the theorem says that p and q must be equal. Hence p = q with p + q = 110 implies that 2p or 2q = 110. and thus p = q = 55 as before.
Method 4. [ Squares are non-negative ]
We all “know” that numbers or algebraic expressions that are squares are “non-negative”. Hence we “choose” appropriately with some “ingenuity” an expression that can help us…
Consider, ( p – q) ^2 which is a square and is non-negative.
[ p – q ] ^2 >= 0
[ p + q ] ^2 – 4pq >= 0 because [ p – q ] ^2 = [ p + q ] ^2 – 4pq;
[ p + q ] ^2 >= 4pq
[ (p + q)/2 ] ^2 >= pq
55^2 >= pq because given that p + q = 110.
Hence for max(pq) = 55, we have pq = 55 and p + q =110. Two equations and two unknowns. We solve simultaneously to get p = q = 55.
There is other methods such as using the “Cauchy Schwarz” inequality etc… but I guess the above suffices. You may observe that a lot of the method above are quite related to each other.
The reason I gave so many “different ways” is so that one can see that in Mathematics, there is often more than one solution and also that different idea or way of looking at things can lead to many beautiful solutions.
