A carpenter is building a rectangular room with a fixed perimeter of 112 ft. What dimensions would yield the maximum area? What is the maximum area?
If you show the work you have done thus far, and explain specifically what you do not understand, perhaps we can help.
Regards,
Rich B.
how to find the answer
Smoke weed every day.
To find the dimensions that yield the maximum area, we can use the concept of optimization.
Let's assume the length of the room is represented by 'l' and the width is represented by 'w'.
The formula for the perimeter of a rectangle is given by P = 2l + 2w, where P is the fixed perimeter of 112 ft. Using this information, we can express 'w' as a function of 'l' as follows:
112 = 2l + 2w
Rearranging the equation, we get:
w = (112 - 2l)/2
w = 56 - l
To find the area of the room, we use the formula A = l * w. Substituting the value of 'w' obtained above into the formula, we have:
A = l * (56 - l)
A = 56l - l^2
This equation represents a quadratic function. Since the coefficient of the quadratic term is negative (-1 in this case), the graph of the function will be a downward-opening parabola, which means the maximum area will occur at the vertex of this quadratic function.
To find the maximum area, we need to find the x-coordinate of the vertex:
The x-coordinate of the vertex of a quadratic equation in the form of Ax^2 + Bx + C can be found using the formula x = -B / (2A).
In our case, A = -1 and B = 56, so substituting these values into the formula, we get:
l = -56 / (2 * -1)
l = 56/2
l = 28
Therefore, the length of the room for which the area is maximized is 28 ft. To find the corresponding width, we substitute this value of length back into the equation we derived for 'w':
w = 56 - l
w = 56 - 28
w = 28
Now we have determined that the dimensions that yield the maximum area are a length of 28 ft and a width of 28 ft.
To find the maximum area, we substitute these dimensions into the area formula:
A = l * w
A = 28 * 28
A = 784
Therefore, the maximum area of the rectangular room is 784 square feet.