Question

At a certain vineyard it is found that each grape vine produces about 10 lb of grapes in a season when about 700 vines are planted per acre. For each additional vine that is planted, the production of each vine decreases by about 1 percent. So the number of pounds of grapes produced per acre is modeled by A(n) = (700 + n)(10 − 0.01n) where n is the number of additional vines planted. Find the number of vines that should be planted to maximize grape production.

Answer 1

Answer:

Number of vines that should be planted are 150.

Step-by-step explanation:

The number of pounds of pounds of grapes produced per acre is represented by the expression $A_{n}=(700+n)(10-0.01n)$

Where n = additional vines planted

To maximize the production of grapes we will find the derivative of A(n) and equate it to zero.

$A_{n}=(7000+3n-0.01n^{2} )$

$A'_{n}=(3-0.02n)$

For $A'_{n}=0$

3 - 0.02n = 0

0.02n = 3

n = $\frac{3}{0.02}$

n = 150

To check whether the maximum value of the function is at n = 150, we will find the second derivative A(n).

$A''_{n}=-0.02$

Which shows A"(n) < 0

Therefore, A(n) has the maximum value at n = 150.

Therefore, number of vines that should be planted are 150.