2.4 Quadratic Models

A farmer has 3000 yards of fence to enclose a rectangular field A farmer has 3000 yards of fence to enclose a rectangular field. What are the dimensions of the rectangle that encloses the most area? w x The available fence represents the perimeter of the rectangle. If x is the length and w the width , then 2x + 2w = 3000

The area of a rectangle is represented by A = xw Let us express one of the variables from the perimeter equation. 2x + 2w = 3000 x = (3000-2w)/2 x =1500 - w Substitute this into the area equation and maximize for w. A = (1500-w)w = -w2 + 1500w The equation represents a parabola that opens down, so it has a maximum at its vertex point.

The vertex is w = -1500/(-2) = 750. Thus the width should be 750 yards and the length is then x =1500 - 750 = 750 The largest area field is the one with equal sides of length 750 yards and total area: A = 7502=562,500 sq.y.

A projectile is fired from a cliff 400 feet above the water at an inclination of 45’ to the horizontal, with a given muzzle velocity of 350ft per second. The height of the projectile above water is given by the equation below where x represents the horizontal distance of the projectile from the base of the cliff. Find the maximum height of the projectile.

To find the maximum height we need to find the coordinates of the vertex of the parabola that is represented by the above equation.

The maximum height is 1357.03 feet and the projectile reaches it at 1914.0625 feet from the base of the cliff.

Vertex (1914.0625, 1357.03 ) (0,400)