Question

a. if the area of a rectangle remains constant but its perimeter increases, how has the shape of the rectangle changed?
b. if the perimeter of a rectangle remains constant but its area increases, how does the shape of the rectangle change?
a. choose the correct answer below.
a. the shape hasnt changed.
b. the difference between the sides is getting bigger.
c. it is not possible for the area of the rectangle to remain constant while the perimeter changes.
d. the difference between the sides is getting smaller.
b. choose the correct answer below.
a. the shape hasnt changed.
b. the difference between the sides is getting bigger.
c. the difference between the sides is getting smaller.
d. it is not possible for the perimeter of the rectangle to remain constant while the area changes.

Explanation:

a. For a rectangle with constant area \(A = lw\) (where \(l\) is length and \(w\) is width), the perimeter \(P=2(l + w)\). Using the AM - GM inequality \(\frac{l + w}{2}\geq\sqrt{lw}\), equality holds when \(l = w\) (a square). If \(A\) is constant and \(P\) increases, \(l + w\) increases. Let \(l=x + k\) and \(w=x - k\) (\(k\geq0\)), \(A=(x + k)(x - k)=x^{2}-k^{2}\). As \(P = 2((x + k)+(x - k))=4x\) (not relevant here, but for fixed \(A=x^{2}-k^{2}\), as \(k\) increases (difference between sides \(2k\) increases), for example, if \(A = 12\), \((l = 12,w = 1)\) (\(P=2(12 + 1)=26\)), \((l = 6,w = 2)\) (\(P=2(6 + 2)=16\)), \((l = 4,w = 3)\) (\(P=2(4+3)=14\)).
b. For a rectangle with constant perimeter \(P = 2(l + w)\) (so \(l + w=\frac{P}{2}\) is constant), the area \(A=lw\). Using \(A=l(\frac{P}{2}-l)=\frac{P}{2}l-l^{2}\), this is a quadratic function \(y=-x^{2}+\frac{P}{2}x\) (where \(x = l\)), and its maximum occurs at \(x=\frac{P}{4}\) (when \(l = w\), a square). As the area increases, the rectangle becomes more like a square, so the difference between the sides (\(l - w\)) gets smaller.

Answer:

a. B. The difference between the sides is getting bigger
b. C. The difference between the sides is getting smaller