Find the length and width of a rectangle with maximum area that has a perimeter of (7P) units.

units (smaller value)

units (larger value)

don't know why theres two dislikes, Just did his method and hes right. Good stuff bob.

Why did the rectangle go on a diet? Because it wanted to maximize its area with the least amount of perimeter!

The perimeter of a rectangle is given by the formula: P = 2(l + w), where P is the perimeter, l is the length, and w is the width.

In this case, we have a perimeter of 7P units, so we can rewrite the formula as: 7P = 2(l + w).

Now, let's solve for one of the variables in terms of the other to calculate the maximum area. But before we do that, let me quickly grab my calculator... *rummages around* Ah, here it is!

Okay, let's continue. Solving the equation for l, we get: l = (7P - 2w) / 2.

Now, we can substitute this value of l into the formula for the area of a rectangle. The area, A, is given by the formula: A = lw.

Substituting the value of l, we get: A = (7P - 2w)w.

To find the maximum area, we need to maximize this function by finding its critical points. However, since I'm an expert in humor, not calculus, let's turn this into a joke instead:

Why did the rectangle hire a coach to find its maximum area? Because it needed someone who could calculate the critical points without getting emotionally invested!

Alright, onto finding the critical points. To do this, we can take the derivative of the area function with respect to w, set it to zero, and solve for w. But wait, let me double-check that I'm not confusing calculus with calisthenics... *checks notes* Nope, this is definitely calculus. Crisis averted!

After some calculations, we find that the width, w, that maximizes the area is given by w = 7P / 6. Substituting this value back into the equation for l, we get l = (7P - 2(7P/6)) / 2.

Simplifying this expression, we find that l = P / 3.

So, the length is P / 3 units and the width is 7P / 6 units.

But hey, don't take my word for it. It's always a good idea to double-check these mathy things with a calculator or a math wizard.

To find the length and width of a rectangle with maximum area that has a perimeter of (7P) units, we can use the fact that for a given perimeter, the rectangle with the maximum area is a square.

Let's first find the side length of the square.

Given that the perimeter is (7P) units and the square has four equal sides, we can set up the equation:

4 * side length = 7P

Simplifying the equation:

side length = (7P) / 4

To find the length and width of the rectangle, we can use the side length of the square.

Since the rectangle is not a square, we know that the length is larger than the width.

So, the width of the rectangle will be (7P) / 4 units (smaller value), and the length will be (7P) / 4 units (larger value).

To find the length and width of a rectangle with maximum area given a perimeter of (7P) units, we can use the concept of calculus. The problem can be solved by finding the critical points of the area function and determining if any of them correspond to a maximum.

Let's proceed with the solution:

1. Let the length of the rectangle be denoted as L and the width as W.
2. The perimeter of a rectangle is given by the formula: 2L + 2W. Since the given perimeter is (7P), we can write the equation as: 2L + 2W = 7P.
3. Rearrange the equation to express one of the variables in terms of the other. Let's solve it for W:
2W = 7P - 2L
W = (7P - 2L) / 2
4. The area (A) of a rectangle is given by the formula: A = L * W.
5. Substitute the expression for W from step 3 into the area formula:
A = L * ((7P - 2L) / 2)
Simplifying, we get: A = (7PL - 2L²) / 2.
6. To find the maximum area, we need to find the critical points. Take the derivative of the area function with respect to L:
dA/dL = 7P/2 - 2L
7. Set the derivative equal to zero to find the critical points:
7P/2 - 2L = 0
7P/2 = 2L
L = (7P/2) / 2
L = 7P/4
8. Substitute L = 7P/4 back into the equation from step 3 to find the corresponding width:
W = (7P - 2(7P/4)) / 2
Simplifying, we get: W = P/4.
9. Therefore, the length is 7P/4 units, and the width is P/4 units.

So, the length of the rectangle is (7P/4) units, and the width is (P/4) units.

area=Lw

perimeter=2L+2W

area=LW
7P=2L+2W
L=7P/2-W

area=W(7P/2-W)
d area/dw=0=7P/2-2W

W=7P/4
then L=7P/2-7P/4=7P/4

areamax=Lw=7P/4 ^2=49P^2/16