The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 22 units long. (a) Express the y-coordinate of P in terms of x. (Hint: Write an equation for the line AB.) (b) Express the area of the rectangle in terms of x. (c) What is the largest area the rectangle can have, and what are its dimensions? P(x,?) -11 11
The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 22 units long. (a) Express the y-coordinate of P in terms of x. (Hint: Write an equation for the line AB.) (b) Express the area of the rectangle in terms of x. (c) What is the largest area the rectangle can have, and what are its dimensions? P(x,?) -11 11
Calculus: Early Transcendentals
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Chapter1: Functions And Models
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![The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 22 units long.
**(a)** Express the y-coordinate of P in terms of x. (Hint: Write an equation for the line AB.)
**(b)** Express the area of the rectangle in terms of x.
**(c)** What is the largest area the rectangle can have, and what are its dimensions?
### Diagram Explanation:
- The diagram illustrates an isosceles right triangle oriented with its base on the x-axis and its vertex B on the y-axis. The triangle's hypotenuse (AB) is 22 units long, spanning from point A on the x-axis at (11, 0) to B on the y-axis at (0, 11).
- A rectangle is inscribed within the triangle. The rectangle has one vertex on the x-axis at (-x, 0) and the opposite vertex at P(x, ?). The other two vertices lie on the line segment AB and the x-axis respectively.
- The point P has coordinates (x, y) where the line AB determines y in terms of x.
- Line AB forms a right angle with the x and y axes at B. Therefore, the slope of AB is determined by the rise over run, which is -1, given it is an isosceles right triangle.
- Since AB forms a line, its equation can be expressed in slope-intercept form \(y = -x + 11\).
**Solution Steps:**
- To find the y-coordinate of point P, use the line equation: \(y = -x + 11\).
- For part **(b)**, the dimensions of the rectangle are determined by its width \(2x\) (spanning from -x to x) and height \(y = -x + 11\). Therefore, the area \(A\) can be expressed as:
\[ A = 2x(y) = 2x(-x + 11) = 22x - 2x^2 \]
- For part **(c)**, to maximize the area of the rectangle, take the derivative of the area \(A\) with respect to \(x\), set the derivative equal to zero, and solve for \(x\). These calculations would yield the dimensions that give the maximum area.
This setup and analysis are often used in calculus and geometry to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8084badb-1caf-4ac3-abaa-e998705ca4db%2Fe293c400-b300-475b-a19d-404e5ab82d12%2Feiz1ge_processed.png&w=3840&q=75)
Transcribed Image Text:The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 22 units long.
**(a)** Express the y-coordinate of P in terms of x. (Hint: Write an equation for the line AB.)
**(b)** Express the area of the rectangle in terms of x.
**(c)** What is the largest area the rectangle can have, and what are its dimensions?
### Diagram Explanation:
- The diagram illustrates an isosceles right triangle oriented with its base on the x-axis and its vertex B on the y-axis. The triangle's hypotenuse (AB) is 22 units long, spanning from point A on the x-axis at (11, 0) to B on the y-axis at (0, 11).
- A rectangle is inscribed within the triangle. The rectangle has one vertex on the x-axis at (-x, 0) and the opposite vertex at P(x, ?). The other two vertices lie on the line segment AB and the x-axis respectively.
- The point P has coordinates (x, y) where the line AB determines y in terms of x.
- Line AB forms a right angle with the x and y axes at B. Therefore, the slope of AB is determined by the rise over run, which is -1, given it is an isosceles right triangle.
- Since AB forms a line, its equation can be expressed in slope-intercept form \(y = -x + 11\).
**Solution Steps:**
- To find the y-coordinate of point P, use the line equation: \(y = -x + 11\).
- For part **(b)**, the dimensions of the rectangle are determined by its width \(2x\) (spanning from -x to x) and height \(y = -x + 11\). Therefore, the area \(A\) can be expressed as:
\[ A = 2x(y) = 2x(-x + 11) = 22x - 2x^2 \]
- For part **(c)**, to maximize the area of the rectangle, take the derivative of the area \(A\) with respect to \(x\), set the derivative equal to zero, and solve for \(x\). These calculations would yield the dimensions that give the maximum area.
This setup and analysis are often used in calculus and geometry to
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