find two positive integers x and y such that their sum is 18, and the sum of their squares is a minimum. a…

find two positive integers x and y such that their sum is 18, and the sum of their squares is a minimum. a. enter the function you want to minimize: a = b. enter the constraint equation: c. plug the constraint into a(x) and enter the new function: a(x)= d. find the derivative: a(x)= e. find the location where a(x) is minimized: x = f. find the minimum value: question help: video message instructor post to forum submit question

find two positive integers x and y such that their sum is 18, and the sum of their squares is a minimum. a. enter the function you want to minimize: a = b. enter the constraint equation: c. plug the constraint into a(x) and enter the new function: a(x)= d. find the derivative: a(x)= e. find the location where a(x) is minimized: x = f. find the minimum value: question help: video message instructor post to forum submit question

Answer

Explanation:

Step1: Define the function to minimize

We want to minimize the sum of squares of two numbers $x$ and $y$. So $A=x^{2}+y^{2}$.

Step2: Write the constraint equation

Given that $x + y=18$, so $y = 18 - x$. The constraint equation is $x + y=18$.

Step3: Substitute the constraint into the function

Substitute $y = 18 - x$ into $A=x^{2}+y^{2}$, we get $A(x)=x^{2}+(18 - x)^{2}=x^{2}+324-36x+x^{2}=2x^{2}-36x + 324$.

Step4: Find the derivative

Using the power - rule $(x^{n})^\prime=nx^{n - 1}$, $A^\prime(x)=\frac{d}{dx}(2x^{2}-36x + 324)=4x-36$.

Step5: Find the critical point

Set $A^\prime(x)=0$, then $4x-36 = 0$. Solving for $x$ gives $4x=36$, so $x = 9$.

Step6: Find the minimum value

Substitute $x = 9$ into $A(x)$. First, since $y=18 - x$, then $y = 9$. And $A(9)=9^{2}+9^{2}=81 + 81=162$.

Answer:

a. $A=x^{2}+y^{2}$ b. $x + y=18$ c. $A(x)=2x^{2}-36x + 324$ d. $A^\prime(x)=4x-36$ e. $x = 9$ f. $162$