Given the area of a rectangle is A = bh, and assuming the rectangle is open on one side, perimeter b + 2h = 40, what formula will maximize the area of the rectangle
rectangle open on one side ??
Poor wording, I will assume that this is the "classic" problem where you have perimeter on only 3 sides, such as the rectangle is against a building.
b = 40 - 2h
A = bh
= h(40-2h)
= -2h^2 + 40h
dA/dh = -4h + 40
= 0 for a max of A
4h=40
h = 10
b = 40-20 = 20
The maximum area is 10(20) or 200 units^2
To maximize the area of the rectangle, we can use the given equation: perimeter b + 2h = 40.
Let's solve the equation for b:
b = 40 - 2h.
Now substitute the value of b in the equation for the area of a rectangle:
A = bh = (40 - 2h)h = 40h - 2h^2.
To find the maximum value of the area, we need to find the value of h that maximizes the equation 40h - 2h^2.
This is a quadratic equation with a negative coefficient for the term h^2, indicating it's a downward-facing parabola. The maximum value occurs at the vertex of the parabola.
The formula for finding the x-value (or h in this case) of the vertex of a quadratic equation in the form of ax^2 + bx + c is:
x = -b / (2a).
In our case, the quadratic equation is -2h^2 + 40h. So, by using the formula above, we can find the h-value that maximizes the area.
h = -40 / (2*(-2))
h = -40 / (-4)
h = 10.
Therefore, the value of h that maximizes the area of the rectangle is 10.
To find the value of b, we can substitute this value of h into the perimeter equation:
b + 2h = 40
b + 2(10) = 40
b + 20 = 40
b = 20.
The value of b that maximizes the area of the rectangle is 20.
Thus, the formula that maximizes the area of the rectangle is A = bh = 20*10 = 200.
To find the formula that maximizes the area of the rectangle, we can use the given information about the perimeter.
Let's break down the given information:
- The perimeter of the rectangle is given by the formula b + 2h = 40.
- The area of the rectangle is given by the formula A = bh.
We need to determine the formula that will maximize the area of the rectangle while satisfying the given perimeter constraint.
Let's solve the given perimeter equation for b in terms of h, so we can substitute it into the area formula:
b + 2h = 40
b = 40 - 2h
Now, substitute the value of b into the area formula:
A = (40 - 2h)h
To maximize the area, we can differentiate the area formula with respect to h and set it equal to zero. Then solve for h.
dA/dh = 40 - 4h = 0
4h = 40
h = 10
Now substitute the value of h back into the formula for b:
b = 40 - 2(10)
b = 40 - 20
b = 20
Therefore, the formula that maximizes the area of the rectangle is:
A = (40 - 2h)h = (40 - 2(10))10 = 20(10) = 200.