Of course the rectangle of greatest area is a square.
So, if you have 60 cm, each side must be 15 cm
and the largest area is 225 cm^2
A rectangular fence is to be made and the available bar wire is 60cm, what are the dimensions of the rectangular closure that will give the maximum area, what is the maximum area?
3 answers
the rectangle that encloses the maximum area is a square
2 L + 2 W = 60 ... L + W = 30 ... L = 30 - W
A = L * W ... substituting ... A = W (30 - W) = 30 W - W^2
max area is on the axis of symmetry of the parabola
... W = -b / 2 a = -30 / (2 * -1) = 15
substitute back to find L ... (see the square)
2 L + 2 W = 60 ... L + W = 30 ... L = 30 - W
A = L * W ... substituting ... A = W (30 - W) = 30 W - W^2
max area is on the axis of symmetry of the parabola
... W = -b / 2 a = -30 / (2 * -1) = 15
substitute back to find L ... (see the square)
if you're supposed to use calculus
A = 30 W - W^2 ... differentiating ... dA/dW = 30 - 2 W
the slope of the tangent line (1st derivative) is zero at the maximum A
dA/dW = 30 - 2W = 0 ... 30 = 2 W ... W = 15
A = 30 W - W^2 ... differentiating ... dA/dW = 30 - 2 W
the slope of the tangent line (1st derivative) is zero at the maximum A
dA/dW = 30 - 2W = 0 ... 30 = 2 W ... W = 15