a lot owner owns a corner lot , 60mx80m, .find the dimensions x, y of the largest rectangular building he can built pn his lot?
If you didn't want any room outside the building, like a yard, to walk around it, it would be 60 by 80 meters. Is there something you have not included in your post?
To find the dimensions of the largest rectangular building that can be built on a corner lot measuring 60m x 80m, we need to consider the concept of maximizing the area of a rectangle.
The area of a rectangle is calculated by multiplying its length by its width. In this case, we want to maximize the area of the rectangle.
Let's assume x represents the length of the rectangle and y represents the width. To maximize the area, we need to find the values of x and y that would give us the largest product.
Given that the lot owner owns a corner lot with dimensions 60m x 80m, we have:
Length (x) = 60m
Width (y) = 80m
To maximize the area, we can use the following steps:
1. Set up the equation for the area of the rectangle: Area = length * width.
Area = x * y
2. Substitute the given values:
Area = 60 * y = 60y
3. Apply calculus to find the maximum value of the function.
To find the maximum, we take the derivative of the area function with respect to y and set it equal to zero.
d(Area)/dy = 60
Setting it equal to zero: 60 = 0
4. Solve for y:
Since the derivative is constant and not equal to zero, there is no maximum or minimum for the area in this case.
5. Determine the other dimension, x:
Since the derivative is constant, we can choose any value for x.
Therefore, the dimensions of the largest rectangular building that can be built on the corner lot measuring 60m x 80m are determined solely by the choice of the other dimension, x.