Chapter 2, Problem 41RE

To determine

To answer: The questions related to the area of a rectangle whose one vertex is in quadrant I on the graph of $y=10-x^2$ and another at the origin.

Answer to Problem 41RE

Solution:

a. $A=x\left(10-x^2\right)$

Explanation of Solution

Given:

$y=10-x^2$

Calculation:

a. Expression for the area $A$ of a rectangle:

The area of the rectangle is given by,

$A=x\times y$

Substituting the value of $y$, we get,

$=x\left(10-x^2\right)$

$=10x-x^3$

Thus, the area of the rectangle is $A=10x-x^3$.

To determine

To answer: The questions related to the area of a rectangle whose one vertex is in quadrant I on the graph of $y=10-x^2$ and another at the origin.

Answer to Problem 41RE

Solution:

b. $A=12.17$

Explanation of Solution

Given:

$y=10-x^2$

Calculation:

b. Largest area $A$ of the rectangle:

Differentiating the equation of the area and equating to zero, we get,

$\frac{dA}{dx}=10-3x^2$

$10-3x^2=0$

Solving the above equation for $x$, we get,

$x=1.83$

The largest area is obtained by substituting the value of $x$ in the equation for the area.

$A=10\left(1.83\right)-{\left(1.83\right)}^3$

$A=12.17$