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Finding the dimensions of the largest rectangle within a graph

Task:

The task at hand is to find the dimensions of the rectangle of the largest area which can be inscribed in the closed region bounded by the x-axis and graph of y=8-x3. The graph is illustrated below.

Solving:

To figure out the dimensions of the largest rectangle, we first need to find the length and the width of said rectangle. All that is given to us is the original equation of the function, which is y=8-x3 . To start off solving this equation, we need to find out the area of the rectangle. However, we are not given any points on the graph, so we cannot derive the area right away. Area =lw, in this case, l=x and w=y. Since we know that, we can derive the area function, which is y=x(8-x3). This is because:

l=x and w=y & Area =lw

l=x

w=y=8-x3

Therefore:

=lw

=x(8-x3)

Now that we have the area function, we need to figure out x and y. To do this, we must first take the derivative of the area function, and then solve for x. Here is the process:

x8-x3

8x-x4

dydx=8-4x3

As shown, first the x was distributed to8-x3, getting 8x-x4 as a result. Next, the derivative was taken of the function, getting8-4x3. Next, we use the derivative of the function to determine x. This is done by solving:

dydx=8-4x3

8-4x3=0

-4x3=-8

x3=2

x=213

First, what we did was take the derivative of the area function and set it to 0. From then on, we solved for x. To solve, we first added 8 to both sides, cancelling out the 8. We were then left with -4x3=-8. Next, we divided -4 by -8, to get 2. Now the equation looks like, x3=2. Finally, we take the third root of 2, getting, x=213. Now that we have figured out the length, x, or 213 , we can figure out y by plugging 213 back into the original equation.

x=213

y=8-x3

y=8-2133

y=8-2

y=6

To find y, we plugged 213 back into the original equation, y=8-x3. It is simply solved throughout the process above. As a final answer, we get:

Length = 213

Width =...

Task:

The task at hand is to find the dimensions of the rectangle of the largest area which can be inscribed in the closed region bounded by the x-axis and graph of y=8-x3. The graph is illustrated below.

Solving:

To figure out the dimensions of the largest rectangle, we first need to find the length and the width of said rectangle. All that is given to us is the original equation of the function, which is y=8-x3 . To start off solving this equation, we need to find out the area of the rectangle. However, we are not given any points on the graph, so we cannot derive the area right away. Area =lw, in this case, l=x and w=y. Since we know that, we can derive the area function, which is y=x(8-x3). This is because:

l=x and w=y & Area =lw

l=x

w=y=8-x3

Therefore:

=lw

=x(8-x3)

Now that we have the area function, we need to figure out x and y. To do this, we must first take the derivative of the area function, and then solve for x. Here is the process:

x8-x3

8x-x4

dydx=8-4x3

As shown, first the x was distributed to8-x3, getting 8x-x4 as a result. Next, the derivative was taken of the function, getting8-4x3. Next, we use the derivative of the function to determine x. This is done by solving:

dydx=8-4x3

8-4x3=0

-4x3=-8

x3=2

x=213

First, what we did was take the derivative of the area function and set it to 0. From then on, we solved for x. To solve, we first added 8 to both sides, cancelling out the 8. We were then left with -4x3=-8. Next, we divided -4 by -8, to get 2. Now the equation looks like, x3=2. Finally, we take the third root of 2, getting, x=213. Now that we have figured out the length, x, or 213 , we can figure out y by plugging 213 back into the original equation.

x=213

y=8-x3

y=8-2133

y=8-2

y=6

To find y, we plugged 213 back into the original equation, y=8-x3. It is simply solved throughout the process above. As a final answer, we get:

Length = 213

Width =...

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