Answer:
$6.00
Explanation:
Given data
quantity demanded ( x ) ∝ 1 / p^3 for p > 1
when p = $10/unit , x = 64
initial cost = $140, cost per unit = $4
<u>Determine the price that will yield a maximum profit </u>
x = k/p^3 ----- ( 1 ). when x = 64 , p = $10 , k = constant
64 = k/10^3
k = 64 * ( 10^3 )
= 64000
back to equation 1
x = 64000 / p^3
∴ p = 40 / ∛x
next calculate the value of revenue generated
Revenue(Rx) = P(price ) * x ( quantity )
= 40 / ∛x * x = 40 x^2/3
next calculate Total cost of product
C(x) = 140 + 4x
Maximum Profit generated = R(x) - C(x) = 0
= 40x^2/3 - 140 + 4x = 0
= 40(2/3) x^(2/3 -1) - 0 - 4 = 0
∴ ∛x = 20/3 ∴ x = (20/3 ) ^3 = 296
profit is maximum at x(quantity demanded ) = 296 units
hence the price that will yield a maximum profit
P = 40 / ∛x
= ( 40 / (20/3) ) = $6
Answer:
0.76%
Explanation:
Firstly we write out the production function to be
Y = K^0.34L^0.42.
So if we have inputs that are increased by 1%, we will now have a new production function which is
Y = (K + 0.01 of K)^0.34 (L + 0.01 of L)^0.42
We write this in terms of growth rate
The Growth rate of Y = 0.34 x the growth rate of K + 0.42 x the growth rate of L
This gives us the Growth rate of Y = 0.34 x 1% + 0.42 x 1%
= 0.34+0.42
= 0.76%
Answer:
The answer is C) 1.25
Explanation:
Operating Leverage= (operating income + fixed expenses) / operating income
Operating Leverage= ($7,200 + $1,800) / $7,200= 1.25
Answer:
$1,194.05
Explanation:
The applicable formula is A = P x ( 1+ r) ^ n
Where A is the future amount
P is principal amount $1000
r is 6% per year or 0.06
n= time in years; 3 years
Since interest is compounded semi-annually, r will be 0.06 /2 = 0.03
n will be 3 years /2 = 6 periods
A = $1000 x ( 1 + 0.03) ^ 6
A = $1000 x 1.194052
A=$1,194.05
