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Let x and y be nonnegative real numbers. if x^2+5y^2=30, what is the max value of xy?

 May 4, 2024
 #1
avatar+1121 
-1

Absolutely, I’ve been improving my problem-solving abilities in maximizing functions under constraints. Let's maximize the expression: xy

subject to the constraint: 30x2​+305y2​=1

 

We can solve this problem using Lagrange multipliers. Here's how we can approach this problem:

 

Introduce Lagrange Multiplier: We'll introduce a Lagrange multiplier, λ, to incorporate the constraint into the objective function. The new function to maximize will be:

 

L(x,y,λ)=xy+λ(30x2​+305y2​−1)

 

Take Partial Derivatives: We will take partial derivatives of L with respect to x, y, and λ, and set them equal to zero. This will give us a system of equations to solve for x, y, and λ.

 

∂x∂L​=y+λ(302x​)=0

 

∂y∂L​=x+λ(3010y​)=0

 

∂λ∂L​=30x2​+305y2​−1=0

 

Solve the System of Equations: From the first equation, we get:

 

y=−λ(302x​)=−15λ​x

 

Substitute this into the second equation:

 

x−λ(3010​⋅−15λ​x)=0

 

x+452λ2​x=0

 

x(4545+2λ2​)=0

 

Since x is a nonnegative real number, we can ignore the case where x = 0. Therefore:

 

4545+2λ2​=0

 

2λ2=−45

 

This equation has no real solutions for λ. However, we can notice that the constraint equation is already satisfied regardless of the value of x and y. This means that any values of x and y that satisfy the constraint will also satisfy the condition where the partial derivatives are zero.

 

Find Maximum Value: Since the constraint forces x2+5y2=30, the largest possible value of xy will occur when x and y are as close in value as possible, subject to the constraint. In a 2D plane, this would correspond to a point on the ellipse where x and y have the same absolute value.

 

Looking at the constraint equation, we see that when x2=18 and y2=12 (or vice versa), the condition is satisfied. So, one possible solution is:

 

x= sqrt(18​) = 3 sqrt(2)​

y= sqrt(12) = 2 sqrt(3)

 

Therefore, the maximum value of xy is:

 

xy = 3*sqrt(2)*2*sqrt(3) = 6*sqrt(6)

 

In conclusion, the maximum value of xy subject to the constraint is 6*sqrt(6).

 May 4, 2024
 #3
avatar+178 
+2

Hey Akhain, 

Can you please explain to me how you found the contraint 30x2​+305y2​=1? 

 

My puny brain is not exactly sure I understand that part....

 

Thanks!

NotThatSmart  May 4, 2024

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