Building Quadratic Models from Verbal Descriptions and Data Section 4.4 Building Quadratic Models from Verbal Descriptions and Data

OBJECTIVE 1

The marketing department at Texas Instruments has found that, when certain calculators are sold at a price of p dollars per unit, the number x of calculators sold is given by the demand equation (a) R(p) = xp = (26,000 – 160p)p = -160p2 + 26,000p Since x ≥ 0 we have 26,000-160p ≥ 0  p ≤ 162.50. So 0 ≤ p ≤ 162.50. (c) Find vertex: (d) R(81.25) = $1,056,250

x = 26,000 - 160(81.25) = 13,000 (f)

800 2x + 2w = 800 A = xw 2x + 2w = 800 2w = 800 – 2x w = 400 - x A = xw = x(400 – x) = -x2 + 400x x = -b/2a = -400/2(-1) = 200 w = 400 – x = 400 – 200 = 200

Find the maximum height of the projectile. = -.0002x2 + x + 500 (a) Find the maximum height of the projectile. How far from the base of the cliff will the projectile strike the water? h(2500) = -.0002(2500)2 + 2500 + 500 = 1750 ft

(b) When it hits the water the height h = 0. h(x) = -.0002x2 + x + 500 = 0 a = -.0002, b = 1, c = 500

Note: Use 0 and 6000 for min and max for x and use -300 and 2000 for min and max for y. Use scale of 1000 for both.

y = .0001193a2 Using vertex (0, 0) we get y = ax2 Using point (2100, 526) we get y = .0001193a2

OBJECTIVE 2

bushels