Considering all rectangles with a given 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.
Considering all rectangles with a given 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.
Considering all rectangles with the same area, the square results in the smallest perimeter for a given area.
Considering all rectangles with a given perimeter, one side being another straight boundry, the 3 sided
rectangle enclosing the greatest area has a length to width ratio of 2:1
i am having serious optimization problems. i don't get it!!! plz help.
a 216-m^2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. what dimensions for the outer rectangle will require the smallest total length of fence? how much fence will be needed?
you are designing a 1000-cm^3 right circular cylindrical can whose manufacture will take waste into account. tehre is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. the total amount of aluminum used up by the can will therefore be
A = 8r^2 + 2(pi)rh
rather than the A = 2(pi)r^2 + 2(pi)rh in Example 4. In example 4 the ratio of h to r for the most economical can was 2 to 1. what is the ratio now?
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