Question

1. suppose ( s = x + 2y ) is an objective function subject to the constraint ( xy = 50 ), for ( x > 0 ) and ( y > 0 ).

a. eliminate the variable ( y ) from the objective function so that ( s ) is expressed as a function of one variable ( x ).
b. find the absolute minimum value of ( s ) subject to the given constraint.

Explanation:

Step1: Eliminate variable \( y \)

Given \( xy = 50 \), we can solve for \( y \) as \( y=\frac{50}{x} \).
Substitute \( y=\frac{50}{x} \) into \( S=x + 2y \):
\( S(x)=x+2\times\frac{50}{x}=x+\frac{100}{x},x>0 \)

Step2: Find the derivative of \( S(x) \)

Using the power rule, if \( S(x)=x + 100x^{-1} \), then \( S^\prime(x)=1-100x^{-2}=\frac{x^{2}-100}{x^{2}}=\frac{(x - 10)(x + 10)}{x^{2}} \)

Step3: Find critical points

Set \( S^\prime(x)=0 \), so \( \frac{(x - 10)(x + 10)}{x^{2}}=0 \).
Since \( x>0 \), we consider \( x - 10=0 \), which gives \( x = 10 \)

Step4: Use the second - derivative test

Find the second - derivative \( S^{\prime\prime}(x)=200x^{-3}=\frac{200}{x^{3}} \)
When \( x = 10 \), \( S^{\prime\prime}(10)=\frac{200}{10^{3}}=\frac{1}{5}>0 \)
So \( S(x) \) has a local minimum at \( x = 10 \)

Step5: Calculate the minimum value of \( S \)

Substitute \( x = 10 \) into \( S(x) \):
\( S(10)=10+\frac{100}{10}=10 + 10=20 \)

Answer:

a. \( S(x)=x+\frac{100}{x},x>0 \)
b. The absolute minimum value of \( S \) is \( 20 \)