Solve this equation?

Question

The demand for grass seed (in thousands of pounds) at price p dollars is given by the following function.

D(p)= -3p^3-2p^2+1489

Use the differential to approximate the changes in demand for the following changes in p.

a. $4 to $4.16 b, $5 to $5.15

a. The change in demand based on a change in p from $4 to $4.16 is about ____ thousand pounds.

b. The change in demand based on a change in p from $5 to $5.15 is about ____ thousand pounds.

Answer 1

Josh F's answer looks good

Just sort of a check on it is to use the cubic demand curve and plug in 4, 4.16, 5 and 5.15

to get 25.4 and 37.8 instead of 25.6 and 36.75

D(4) = 224

D(4.16) =250.6

250.6 - 224 = 25.4

D(5) = 425

D(5.15) = 462.8

462.8-425 = 37.8

If you use the differential, you extrapolate an instantaneous change to an interval

IF you stick to the cubic equation, you get the actual change over the interval

for small changes, they're both very close

IF this happens to be an economics course, it's necessary to distinguish "Demand" from

"Quantity Demanded." They take points off if you mix them up. A hyper technical

fetish of economics instructors. Or some business instructors or business calculus

instructors.

The cubic equation gives the quantity demanded. The "Demand" doesn't change.

the quantity demanded does change.

Answer 2

ΔD(p) ≈ dD(p) = D'(p)Δp = - (9p² + 4p)Δp;

a. If price changes from $4 to $4.16, then Δp = 4.16 - 4 = $0.16

and ΔD(4) ≈ - (9·4² + 4·4)·0.16 = - 25.6 thusands of pounds

b. If price changes from $5 to $5.15, then Δp = 5.15- 5 = $0.15

and ΔD(5) ≈ - (9·5² + 4·5)·0.15 = - 36.75 thusands of pounds

Answer 3

dD/dp = -9p²-4p , dD/dp |_p=4 = -160 , dD/dp |_{p =5} = -245

The change in demand from $4 -> $4.16 ~ .16(-160) = - 25.6 thousand lbs.

The change in demand from $5 -> $5.15 ~ .15(-245) = - 36.75 thousand lbs