Express the area A of a rectangle with perimeter 100 feet as a function of the l

Question

Express the area A of a rectangle with perimeter 100 feet as a function of the length v of one of its sides. V is the length of one of its sides.
the length of the other side=100-v
V(100-V)=Area +100V=Area

1e. Simplify this function so that it reads as a quadratic equation. +100V=Area


Q12. If you were to sketch the graph of the function in 1e, where the x-axis is labeled as your length (in feet) and the y-axis is labeled as your area (in square feet), you will see that it has the shape of a parabola.

• Use the graph to determine the dimensions of a rectangle with perimeter 100 feet and the following two areas:

• 400 square feet & 500 square feet y=400, x=9.582 and y=500, x=9.472

I am having trouble completing the assignment.
• Explain how 700 square feet is not a possible area in this situation.

3. At what length and width will you have a maximum area?
4. What is your observation from #3? Is there a conclusion that can be made from this observation?

Explanation / Answer

If v is one side, then 50-v is the other, since perimeter is 2l +2w. (2v+2(50-v)= 100) So the area A= 50v- v^2 (lw) If A= 400 we have 50 v-v^2= 400 or v^2- 50v +400= 0 Solving this we have v=40 and 50-v= 10. If A= 500 we have v^2 -50 v +500= 0 v=36.18 and 50-v= 13.82 700 is not a possible solution because the quadratic equation v^2-50v +700 has no real roots. To find the maximum area we take the derivative of the area function A= 50v-v^2 and set it equal to 0. dA/dv= 50- 2v 2v=50 v=25. Max area is where you have a square with side 25. The conclusion that can be drawn is when you have a given perimeter, the maximum area will always be where all sides are equal, in other words a square.

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