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Let P(r, y) be a point on the graph of y = - I + 13 with 0 < z < 13. Let PQRS be a rectangle with one side on the z-axis and two vertices on the graph, as shown on the picture below. Find the rectangle with the greatest possible Area. Round final answer to at least 3 decimal places. Greatest Area: unit? when 2 = and y = R

Let P(r, y) be a point on the graph of y = - I + 13 with 0 < z < 13. Let PQRS be a rectangle with one side on the z-axis and two vertices on the graph, as shown on the picture below. Find the rectangle with the greatest possible Area. Round final answer to at least 3 decimal places. Greatest Area: unit? when 2 = and y = R

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Problem Statement** Let \( P(x, y) \) be a point on the graph of \( y = -x^2 + 13 \) with \( 0 < x < \sqrt{13} \). Let \( PQRS \) be a rectangle with one side on the \( x \)-axis and two vertices on the graph, as shown in the picture below. Find the rectangle with the greatest possible area. Round your final answer to at least 3 decimal places. - Greatest Area: \(\_\_\_\_\_\_\_\_\_\) unit\(^2\) - when \( x = \_\_\_\_\_\_\_\_ \) and \( y = \_\_\_\_\_\_\_\_ \) **Diagram Explanation** The diagram shows the graph of the function \( y = -x^2 + 13 \), which is a downward-opening parabola. The rectangle \( PQRS \) is inscribed such that: - \( PQ \) and \( PS \) are the vertical sides, with the endpoints, \( Q \) and \( P \), lying on the parabola. - \( RS \) is on the \( x \)-axis. - The width of the rectangle lies along the \( x \)-axis from point \( R \) to point \( S \). **Instructions** Use the methods of this class; DO NOT USE CALCULUS. --- For further guidance, students should find the expression for the area of rectangle \( PQRS \) and use algebraic manipulation to solve for the maximum area, adhering to the restriction \( 0 < x < \sqrt{13} \).
Transcribed Image Text:**Problem Statement** Let \( P(x, y) \) be a point on the graph of \( y = -x^2 + 13 \) with \( 0 < x < \sqrt{13} \). Let \( PQRS \) be a rectangle with one side on the \( x \)-axis and two vertices on the graph, as shown in the picture below. Find the rectangle with the greatest possible area. Round your final answer to at least 3 decimal places. - Greatest Area: \(\_\_\_\_\_\_\_\_\_\) unit\(^2\) - when \( x = \_\_\_\_\_\_\_\_ \) and \( y = \_\_\_\_\_\_\_\_ \) **Diagram Explanation** The diagram shows the graph of the function \( y = -x^2 + 13 \), which is a downward-opening parabola. The rectangle \( PQRS \) is inscribed such that: - \( PQ \) and \( PS \) are the vertical sides, with the endpoints, \( Q \) and \( P \), lying on the parabola. - \( RS \) is on the \( x \)-axis. - The width of the rectangle lies along the \( x \)-axis from point \( R \) to point \( S \). **Instructions** Use the methods of this class; DO NOT USE CALCULUS. --- For further guidance, students should find the expression for the area of rectangle \( PQRS \) and use algebraic manipulation to solve for the maximum area, adhering to the restriction \( 0 < x < \sqrt{13} \).
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Let P(z,y) be a point on the graph of y = ²+13 with 0<<√13. Let PQRS be a rectangle with one side on the z-axis and two vertices on the given graph

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Find the rectangle with the greatest possible Area.

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