Derieving Derivative #8

Calculus Level 2

Find the two positive numbers x and y whose sum is 35 and the product x 5 y 2 x^5y^2 is maximum

x = 15 ,y = 20 x = 25 ,y = 10 x = 28 ,y = 7 x = 21 ,y = 14

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3 solutions

Satvik Golechha
Sep 12, 2014

So, we want to maximize the value of x 5 y 2 x^5y^2 . We will definitely use the AM-GM inequality. A first thought would have been to directly apply the simple inequality to the numbers x , x , x , x , x , y , , y {x, x, x, x, x, y, ,y} , but then we will have 5 x + 2 y 5x+2y in the LHS, and who's gonna give us that?

So we need some other set of numbers. Now, we're given that x + y = 35 x+y=35 , so how can we get x + y x+y by a combination of 5 5 x's and 2 2 y's?

For that, we'll apply the AM-GM inequality to x 5 , x 5 , x 5 , x 5 , x 5 , y 2 , y 2 \frac{x}{5}, \frac{x}{5},\frac{x}{5},\frac{x}{5},\frac{x}{5},\frac{y}{2},\frac{y}{2} This will give us x + y x+y , and x 5 y 2 x^5y^2

But wait! Is it the only possible arrangement? We can use it for the set x 10 , x 10 , 6 x 10 , x 10 , x 10 , 3 y 4 , y 4 \frac{x}{10}, \frac{x}{10},\frac{6x}{10},\frac{x}{10},\frac{x}{10},\frac{3y}{4},\frac{y}{4} This will also give us x + y x+y , and x 5 y 2 x^5y^2 .

What's the problem in this? The problem is the "equality" case.

In the AM-GM inequality, we attain equality when all the numbers in the set are equal. But in this set, we can't possibly have 6 x 10 = x 10 \frac{6x}{10}=\frac{x}{10} , (x and y are positive)

So, back to the problem, we have the set. Applying the AM-GM inequality, we get:- x 5 + x 5 + x 5 + x 5 + x 5 + y 2 + y 2 7 ( x 5 y 2 5 5 2 2 ) 1 7 \frac{\frac{x}{5}+ \frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{y}{2}+\frac{y}{2}}{7} \geq (\frac{x^5y^2}{5^52^2})^\frac{1}{7} , putting x + y = 35 x+y=35 , and taking the equality case we get:-

x 5 y 2 = 4 × 1 0 2 x^5y^2=4 \times 10^2 , 2 x = 5 y 2x=5y , x + y = 35 x+y=35 .

Solving them, we get x = 25 x=25 and y = 10 y=10 .

-_- Q.E.D.

Great application of AM-GM. I used calculus for this and I too am feeling somewhat ashamed that I took the easy way out. Keep up the good work and out of the box thinking!! PS -I hope this appreciating comment will not be considered spam.

Shabarish Ch - 6 years, 9 months ago

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Thank s s .

Satvik Golechha - 6 years, 9 months ago

Great solution, Satvik. I used Lagrange multipliers, .i.e., calculus, which was quick in this case but felt like cheating after reading your application of the AM-GM inequality.

Brian Charlesworth - 6 years, 9 months ago

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@brian charlesworth Thank you sir. I don't know any calculus at all. Will you please tell me a good source from where I can learn it from the very basics.

Satvik Golechha - 6 years, 9 months ago

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I still refer back to my textbooks, (my introductory text was by Edwards and Penney), but online you could try KhanAcademy to get you started.

Brian Charlesworth - 6 years, 9 months ago
Chew-Seong Cheong
Sep 12, 2014

Good old calculus can solve it too.

x + y = 35 y = 35 x x+y=35\quad \Rightarrow y = 35 -x

f ( x ) = x 5 y 2 = x 5 ( 35 x ) 2 = x 5 ( 1225 70 x + x 2 ) \Rightarrow f(x) = x^5y^2 = x^5(35-x)^2 = x^5(1225 - 70x +x^2)

f ( x ) = x 7 70 x 6 + 1225 x 5 \Rightarrow f(x) = x^7 - 70x^6 + 1225x^5

f ( x ) = 7 x 6 420 x 5 + 6125 x 4 = x 4 ( x 25 ) ( x + 35 ) f'(x) = 7x^6 - 420x^5 + 6125x^4 = x^4(x-25)(x+35)

At turning point, f ( x ) = 0 f'(x) = 0 , then x = 35 , 0 , 25 x = -35, 0, 25 .

The only acceptable value is x = 25 x = 25 .

It can be found that f ( 25 ) < 0 f''(25)<0 .

Therefore, x 5 y 2 x^5y^2 is maximum when ( x , y ) = ( 25 , 10 ) (x,y) = \boxed{(25,10)} .

I have done this problem by this method..

Parag Zode - 6 years, 9 months ago
Brock Brown
Dec 28, 2014

Quick Pythonic solution:

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biggest = 0
x = 35
y = 0
answers = (0, 0)
while x != 0:
    if x**5*y**2 > biggest:
        biggest = x**5*y**2
        answers = x, y
    x -= 1
    y += 1
print answers

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