Expert answers to "Among all pairs of numbers (x,y) such that..."

osi4a2nxk

Answered question

2021-11-17

Among all pairs of numbers (x,y)
such that

2 x + y = 17

, find the pair for which the sum of squares,

x^{2} + y^{2}

is minimum. Write your answers as fractions reduced to lowest terms.

Answer & Explanation

oces3y

Beginner2021-11-18Added 21 answers

Step 1
Given: $2 x + y = 17$ .
$∴ y = 17 - 2 x$ .
The function $x^{2} + y^{2}$ should be minimum.
$y = 17 - 2 x$ substitute in $y^{2}$
$∴ f (x) = x^{2} + {(17 - 2 x)}^{2}$
$= x^{2} + 289 - 68 x + 4 x^{2}$
$= 5 x^{2} - 68 x + 289$ .
Step 2
In order to find critical point, $f^{'} (x) = 0$ .
$f^{'} (x) = 10 x - 68 = 0 \Rightarrow x = \frac{68}{10}$ .
* To find concave up (or) down, f"(x).
$∴ f x) = 10 > 0 ∴$ Concave up.
So At $x = \frac{68}{10}$ , it is minimum.
when $x = \frac{68}{10}$ , $y = 17 - 2 x = 17 - 2 (\frac{68}{10}) = \frac{17}{5}$ .
$∴ f (x) = x^{2} + y^{2}$ is min at $(x, y) = (\frac{68}{10}, \frac{17}{5})$
Reduced to lowest term
$(x, y) = (\frac{34}{5}, \frac{17}{5})$

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