Shondra wants to enclose a rectangular garnden with 240 yards of fencing. What dimensions for the garden will maximize its area?

2W + 2L = 240

Largest area will be a square.

60 * 60 = 3600
70 * 50 = 3500
80 * 40 = 3200

Well, Shondra, let me tell you a not-so-rectangular secret. If you want to maximize the area of a rectangular garden, you have to remember one thing: the dimensions have to be equal! That's right, you need a perfectly square garden, my friend. So, if you divide the 240 yards of fencing equally between all four sides, you'd have 60 yards per side. Now, if you squint your eyes and look at it, the garden will look like a perfect square, with each side measuring 60 yards. Voila! You've achieved maximum gardeniness!

To find the dimensions of the garden that will maximize its area, we can use the concept of optimization.

Let's assume the length of the garden is L yards and the width is W yards. The perimeter of the rectangular garden is given as 240 yards, so we can write the equation:

2L + 2W = 240

To simplify the equation, we can divide both sides by 2:

L + W = 120

From this equation, we can rewrite L in terms of W:

L = 120 - W

Now, let's express the area of the garden in terms of L and W. The area of a rectangle is given by:

A = L × W

Substituting the expression for L from above, we have:

A = (120 - W) × W

Expanding this equation, we get:

A = 120W - W^2

Now, we can find the maximum area of the garden by taking the derivative of A with respect to W and setting it equal to zero. Let's differentiate A:

dA/dW = 120 - 2W

Setting this derivative equal to zero:

120 - 2W = 0

Solving for W:

2W = 120

W = 60

Substituting this value of W back into the equation for L, we get:

L = 120 - 60
L = 60

Therefore, the dimensions for the garden that maximize its area are a length of 60 yards and a width of 60 yards.

To find the dimensions of the garden that will maximize its area, we can use the concept of optimization.

Let's assume the length of the garden is 'L' yards, and the width of the garden is 'W' yards.

We know that the perimeter of a rectangle is given by the formula: P = 2L + 2W, where P represents the perimeter.

In this case, the perimeter of the garden is 240 yards, so we can write the equation as: 240 = 2L + 2W.

To find the dimensions that maximize the area, we need to express the area of the garden in terms of a single variable. The area of a rectangle is given by the formula: A = L * W.

We can rearrange the equation for the perimeter to solve for L: L = (240 - 2W) / 2.

Substituting the value of L into the equation for the area, we get: A = (240 - 2W) / 2 * W.

Simplifying this expression, we have: A = (240W - 2W^2) / 2.

To maximize the area, we need to find the value of W that will make the derivative of the area equation equal to zero. To find this critical point, we can differentiate the area equation with respect to W.

dA/dW = 240/2 - 2W = 120 - 2W.

Setting the derivative equal to zero, we have: 120 - 2W = 0.

Solving for W, we get: W = 60.

Substituting this value of W back into the equation for L, we can find the length of the garden: L = (240 - 2(60)) / 2 = 60.

Therefore, the dimensions of the garden that will maximize its area are 60 yards by 60 yards.