CAN SOMEONE PLEASE ANSWER THIS QUESTION QUICK!!

Question

Let’s say we have access to planting 2 different crops, Watermelons and Grapes. We can find the revenue of a crop by the equation: R = (p0 + px)(n0 - nx) Where R is the revenue, p0 is the starting price, n0 is the amount sold at the starting price, p is the price increase, n is the decrease in the amount sold per price increase, and x is the number of price increases. Crop A: Watermelons have a starting price of $12.00 per m2 of area sold. At this price, you are able to sell 45 m2 worth. For every $2.00 increase, you will sell 1 m2 less. So the first task is to figure out how to put these values into the equation above, and turn it into a quadratic function. Then you will need to find the value of x in order to make the revenue as large as possible. Once you know x, the number of price increases, you can then find what you should price your Watermelons per m2 .

Equation: R =

Number of price increases:x=

Price of Watermelon per m^2=

Answer 1

R=(12+2x)(45-x) answer

R=540+90x-12x-2x^2

R=540+78x-2x^2

this is a parabola that opens downwards so it has a maximum point

this maximum occurs at (h,k), the vertex of the parabola

h=-b/2a

h=-78/-4

h=19.5 increases

x=19.5 increases of $2.00 each answer

2*19.5=$39 increase

$12+$39=$51.00 per m^2 (price of watermelon)

45-19.5=25.5 m^2 sold answer

check:

51*25.5=$1300.50 profit (12+2*19.5 and 45-19.5)

if x=19, you have 50*26=$1300 profit only, $0.50 less

if x=20, you have 52*25=$1300 profit only again

if x=18.50, you have 49*26.5=$1298.50

if x=20.5, you have 53*24.5=$1298.50

Answer 2

Hi Sylvia. I noticed the problem mentioned Grapes but doesnt explain anything about them. Anyway, if I am understanding this correctly, we would substitute the following for Watermelons:

p0=$ 12 per m^2

n0 = 45 ^ 2

p = $ 2

n = 1 m^2

so the equation for revenue of Watermelons would be R = (12+2x)(45 - 1x) = -2x^2+78x+540

The derivative of R would be - 4x+78

Setting this to 0 to find the optimal number of price increases for maximum revenue would yield:

x = 19.5

I'm not clear if there is such a thing as a half of a number of price increases, but I will assume that it is possible.

Placing this value of x into the first portion of the equation of R(x) would yield:

(p0+px) = 12 + 2(19.5) = $ 51.00 per m^2

I hope I understood this problem correctly.