Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region enclosed by the graphs f(x) = 18 - x^(2) and g(x) = 2 x^(…

Question

Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region enclosed by the graphs f(x) = 18 -
`x^(2)` and g(x) = 2
`x^(2)` - 9 by answering the following:

a. Describe the width of the rectangle in terms of x
b. Describe the height of the rectangle in terms of f(x) and g(x)
c. Describe the height of the rectangle in terms of x
d. Write the function A(x) of the area of the rectangle in terms of the width and height found in parts a& c.
e. Find the maximum of A

Answer 1

ιHi

We know that the length of the vertical side of the rectangle is the distance between the two function.

So we have l = f(x) - g(x)

Don't forget that we have two of these sides at two different x values.

We don't want to get confuse so lets call the left side of the y axis x1 and x2 for the right axis. ( when you're writing it put the x bigger than 1 or 2)

So we have L1 = f(x1) -g(x1)

and L2 = f(x2)-g(x2)

Before we start solving, we need to remember that there is no Y value inside the graphs. We have the others sides W, each of them intersect one function twice.

w1 = w2 = x2 - x1

But we know that f and g are symmetric across y axis, for that reason x1= -x2

Hint: Symmetric is when one shape becomes exactly like another if you flip, slide or turn it. (Google definition)

Let's go back to work

Now we have w=-x1 -x1 = -2x1

Now plug L1 and W into the area equation

A = L * W = (f(x1) - g(x1)) * (-2x1)

Now replace their values

A(x) = ((18-x²) - (2x²-9)) * (-2x)

A(x) = -2x (27-3x²)

A(x) = 6x³ -54x

We need to optimize this function. How?

Take the derivative and find the Zeros

A(x) = 18x² -54 = 0

x² = 3

x = +/- √3

Now we have the negative value, which is x

Plug it into the area of function

A(-√3)= 6(-√3)³ -54 (-√3)

= 54√3 - 18√3

= 36√3

= 62.35

This is the final answer.

I hope that's help.. if not I am sorry.

I don't solved by myself, my friends help me too. So that's why it took me so long to answer.