A field is to be fenced in on 3 sides with a river along one side, as shown.Find the maximum possible area [math: A] (in [math: m2]) of the field, given that the fence is to be constructed using 2900 metres of fencing materials.

Question

BlackTom AI · Accepted Answer

The problem describes a rectangular field that is fenced on three sides, with a river along the fourth side, so only two shorter sides and one long side require fencing.
First, denote the side along the river as x (the length of the unfenced side is x) and the two perpendicular sides as y. The total fencing used is along the two y-sides and the far x-side, giving the perimeter constraint: x + 2y = 2900.
The area of the rectangle is A = x*y. To maximize A under the constraint, express y in terms of x: y = (2900 - x)/2, so A(x) = x*(2900 - x)/2 = (2900x - x^2)/2.
To find the maximum, either complete the square or take a derivative: A'(x) = (2900 - 2x)/2 = 1450 - x. Setting A'(x) = 0 yields x = 1450.
Substitute back to find y: y = (2900 - 1450)/2 = 1450/2 = 725.
Now compute the maximum area: A = x*y = 1450 * 725 = 1,051,250.
Thus, the configuration that maximizes the area uses a side along the river of 1450 m and the two fenced sides of 725 m each, giving the maximum area of 1,051,250 m².

A field is to be fenced in on 3 sides with a river along one side, as shown.Find the maximum possible area [math: A] (in [math: m2]) of the field, given that the fence is to be constructed using 2900 metres of fencing materials.数值题

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