Find the Maximum area for the given perimeter of a rectangle. State the length and width of the rectangle.
28 inches
Well, finally a calculus problem.
Ok, we know that the area for a rectangle is
A=l*w and the perimeter is P=2(l+w)
In this problem P= 28, so let's express one of the dimensions in terms of the other an substitute into the area formula. Thus 28=2(l+w). Let's solve for l in terms of w, thus 14=l+w, or l=14-w When we substitute this into the area formula we get
A=(14-w)*w. So A=14w-w^2
Now find dA/dw and evaluate the critical points.
dA/dw = 14-2w and dA/dw = 0 means 14-2w=0
So w = 7 is a critical point. I'll let you verify that this is the max. (the second deriv is -2, what does that mean?)
Thus the rectangle of maximum area has w=7. If you put this back into the formula for the perimeter, you'll find that l=7 too. This means that the rectangle of max. area is a square with a side=P/4.
Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.
This should give you your answer.
i dont have an answer but just pointing out that 256/16 is 16 not 15 so xandy are both 16
To find the maximum area for a given perimeter of 28 inches, we can use the formula for the perimeter of a rectangle, which is P=2(l+w), where l is the length and w is the width.
In this case, the perimeter is given as 28 inches, so we have the equation 28=2(l+w).
To find the maximum area, we need to express one of the dimensions (length or width) in terms of the other. Let's express the length in terms of the width.
From the equation 28=2(l+w), we can solve for l:
l=14-w.
Now, substitute this expression for l in the area formula A=l*w:
A=(14-w)*w.
Simplify the equation to get the area in terms of the width:
A=14w-w^2.
To find the maximum area, we can differentiate the area function with respect to the width (w) and set the derivative equal to 0. This will give us the critical point where the area is maximized.
Differentiating A=14w-w^2 with respect to w, we get:
dA/dw=14-2w.
Set dA/dw=0 and solve for w:
14-2w=0.
2w=14.
w=7.
So, the critical point where the area is maximized is when the width (w) is 7 inches.
To find the length (l), substitute the value of w=7 back into the expression for l:
l=14-w=14-7=7.
Therefore, the rectangle with maximum area for a perimeter of 28 inches has a length of 7 inches and a width of 7 inches.
To find the maximum area for a given perimeter of a rectangle, we can apply calculus by finding the dimensions that maximize the area function.
Let's start with a rectangle with length (l) and width (w). The formula for the perimeter is P = 2(l + w), and the formula for the area is A = l * w.
Given that the perimeter is 28 inches, we have P = 2(l + w) = 28.
To find the maximum area, we need to express one of the dimensions in terms of the other and substitute it into the area formula. Let's solve the perimeter equation for l in terms of w:
2(l + w) = 28
l + w = 14
l = 14 - w
Now, substitute the expression for l into the area formula:
A = l * w
A = (14 - w) * w
A = 14w - w^2
To find the maximum area, we need to find the critical points of the area function. Take the derivative of A with respect to w:
dA/dw = 14 - 2w
Now, set dA/dw equal to zero to find the critical points:
14 - 2w = 0
2w = 14
w = 7
The critical point is w = 7. To determine if it's a maximum or minimum, we can examine the second derivative of A. Taking the derivative of dA/dw:
d^2A/dw^2 = -2
Since the second derivative is negative (-2), this means that w = 7 corresponds to a maximum point.
Therefore, the rectangle of maximum area for a perimeter of 28 inches has width w = 7 inches. By substituting w back into the perimeter equation, we find length l = 14 - w = 14 - 7 = 7 inches.
This means that the rectangle with maximum area is a square with sides measuring 7 inches, as the width and length are the same for the rectangle.