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

6. suppose ( s = x + 2y ) is an objective function subject to the constraint ( xy = 50 ), for ( x > 0 ) and ( y > 0 ).\na. eliminate the variable ( y ) from the objective function so that ( s ) is expressed as a function of one variable ( x ).\nb. find the absolute minimum value of ( s ) subject to the given constraint.

6. suppose ( s = x + 2y ) is an objective function subject to the constraint ( xy = 50 ), for ( x > 0 ) and ( y > 0 ).\na. eliminate the variable ( y ) from the objective function so that ( s ) is expressed as a function of one variable ( x ).\nb. find the absolute minimum value of ( s ) subject to the given constraint.

Answer

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 )