Question

1. suppose the objective function ( p = xy ) is subject to the constraint ( 10x + y = 100 ), where ( x ) and ( y ) are real numbers.

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

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 \( x \):

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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 \( 250 \)