Question

Calculus Problem: A businessman found out that his profit varies as the product of the amount spent for production and the square root of the amount spent for advertising. If his total available budget for these expenses is 1.5 Million, how should he alocate his funds to maximize his profits?

Answer 1

$$\rm profit = k * production * \sqrt{ advertising}\\
production + advertising = 150,000,000

$$

Answer 2

let profit = P , production cost = x , advertising = y.

$$
\Large P= k\cdot x ~ \sqrt{y}\\
\Large x+y = 150,000,000
\\ \therefore \\
\Large x = 150,000,000-y
\\ \therefore \\
\Large P= k~( 150,000,000-y) ~ \sqrt{y}\\
\\ \therefore \
\Large \frac{dP}{dy}=k [~(-1)\sqrt{y} +( 150,000,000-y) \frac{1}{\sqrt{y}}]
$$

Answer 3

it would be easier if you first distributed the y

Answer 4

Should 1.5 million be 1,500,00 not 150,000,000 ?

Answer 5

let profit = P , production cost = x , advertising = y.

$$
\Large P= k\cdot x ~ \sqrt{y}\\
\Large x+y = 150,000,000
\\ \therefore \\
\Large x = 150,000,000-y
\\ \therefore \\
\Large P= k~( 150,000,000-y) ~ \sqrt{y}\\
\\ \therefore \\
\Large P= k~( 150,000,000-y) ~ y^\frac{1}{2}\\
\\ \therefore \\
\Large P= k~( 150,000,000 \cdot y^\frac{1}{2}-y\cdot ~ y^\frac{1}{2} )\\
\\ \therefore \\
\Large P= k~( 150,000,000 \cdot y^\frac{1}{2}- y^\frac{3}{2} )\\
\\ \therefore \\
\Large \frac{dP}{dy}= k (150,000,000 \cdot \frac{1}{2}y^{-\frac{1}{2}} - \frac{3}{2}y^\frac{1}{2})
$$

Answer 6

now solve dP / dy = 0

Answer 7

@ perl \[0=k[(1,500,000)(\frac{ 1 }{ 2 }y^{-\frac{ 1 }{ 2 }})-\frac{1 }{ 2 }y ^{\frac{ 1 }{ 2 }}]\]

Answer 8

@perl what's the variable "k" for?

Answer 9

it says it is proportional, and k is a constant , not a variable

Answer 10

P varies as x and square root y
$$ \\ \therefore\\$$
P = k * x * sqrt(y) , for some constant k

Answer 11

@perl can you explain how to solve dP/dy = 0 ? was my equation above correct?

Answer 12

$$
\Large k (150,000,000 \cdot \frac{1}{2}y^{-\frac{1}{2}} - \frac{3}{2}y^\frac{1}{2})=0\\
\text {divide both sides by k , which we can assume is non zero}
\\ \Large 150,000,000 \cdot \frac{1}{2}y^{-\frac{1}{2}} - \frac{3}{2}y^\frac{1}{2}=0\\
\\ \Large 75,000,000 ~y^{-\frac{1}{2}} - \frac{3}{2}y^\frac{1}{2}=0\\
\\ \Large 75,000,000 ~y^{-\frac{1}{2}} = \frac{3}{2}y^\frac{1}{2}

\\ \Large \frac{2}{3} \cdot75,000,000 ~y^{-\frac{1}{2}} = y^\frac{1}{2}
\\ \Large \frac{2}{3} \cdot75,000,000 ~y^{-\frac{1}{2}} = y^\frac{1}{2}
\\ \Large 50,000,000 ~y^{-\frac{1}{2}} = y^\frac{1}{2}

\\\large \text{multiply both sides by } y^\frac{1}{2} \\\
\\ \Large 50,000,000 ~y^{-\frac{1}{2}}\cdot y^\frac{1}{2} = y^\frac{1}{2}
\cdot y^\frac{1}{2}

\\ \Large 50,000,000 ~y^{-\frac{1}{2}+ \frac{1}{2}} = y^{\frac{1}{2} +\frac{1}{2}}
\\ \Large 50,000,000 ~y^{0} = y^{1}
\\ \Large 50,000,000\cdot 1 = y

\\ \Large 50,000,000 = y

$$

Answer 13

so x therefor is equal to x=150,000,000-50,000,000

x = 100,000,000

Answer 14

@perl is this correct? do i need to find the value of "k"?

Answer 15

you don't need to know the value of k, since we divided it out when we solved dP/dy = 0

the particular value of k does not affect the problem .

Answer 16

ok, thanks for the help!