A truck that is 11 feet tall and 7 feet wide is traveling under an arch. The arch can be modeled by: y=-0.0625x+1.25x +5.75 where x and y are measured in feet.

a) Will the truck fit under the arch? Explain (with numerical details).

BIRD

b) What is the maximum width that a truck 11 feet tall can have and still make it under the arch?

Yes

8 ft

Step-by-step explanation:

The equation of the arch is

Differentiating with respect to we get

Equating with zero

Double derivative of the equation

So, the function is maximum at

The function is maximum at

So, the truck's center should be below the point for maximum width to pass through.

Now truck is 11 feet tall so

The maximum width of the truck can be if it is 11 feet tall.

So, the truck which has a width of 7 feet can fit under the arch.

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt