5. suppose the objective function ( p = xy ) is subject to the constraint ( 10x + y = 100 ), where ( x ) and…

5. suppose the objective function ( p = xy ) is subject to the constraint ( 10x + y = 100 ), where ( x ) and ( y ) are real numbers.\na. eliminate the variable ( y ) from the objective function so that ( p ) is expressed as a function of one variable ( x ).\nb. find the absolute maximum value of ( p ) subject to the given constraint.

5. suppose the objective function ( p = xy ) is subject to the constraint ( 10x + y = 100 ), where ( x ) and ( y ) are real numbers.\na. eliminate the variable ( y ) from the objective function so that ( p ) is expressed as a function of one variable ( x ).\nb. find the absolute maximum value of ( p ) subject to the given constraint.

Answer

Explanation:

Step1: Express ( y ) in terms of ( x )

From the constraint ( 10x + y=100 ), we can solve for ( y ) as ( y = 100 - 10x ).

Step2: Substitute ( y ) into the objective function

Substitute ( y = 100 - 10x ) into ( P=xy ). Then ( P(x)=x(100 - 10x)=100x-10x^{2} ).

Step3: Find the derivative of ( P(x) )

Using the power rule, if ( P(x)=100x - 10x^{2} ), then ( P^{\prime}(x)=\frac{d}{dx}(100x)-\frac{d}{dx}(10x^{2})=100 - 20x ).

Step4: Find the critical points

Set ( P^{\prime}(x)=0 ), so ( 100 - 20x = 0 ). Solving for ( x ): [ \begin{align*} 100-20x&=0\ 20x&=100\ x& = 5 \end{align*} ]

Step5: Find the second - derivative of ( P(x) )

Differentiate ( P^{\prime}(x)=100 - 20x ) with respect to ( x ). ( P^{\prime\prime}(x)=\frac{d}{dx}(100)-\frac{d}{dx}(20x)=- 20). Since ( P^{\prime\prime}(5)=-20<0 ), ( x = 5 ) is a point of maximum.

Step6: Find the value of ( y )

Substitute ( x = 5 ) into ( y=100 - 10x ). Then ( y=100-10\times5 = 50 ).

Step7: Find the maximum value of ( P )

Substitute ( x = 5 ) and ( y = 50 ) into ( P=xy ). Then ( P=5\times50 = 250 ).

Answer:

a. ( P(x)=100x - 10x^{2} ) b. The absolute maximum value of ( P ) is ( 250 )