Question

Hallamos precios justos empleando funciones cuadráticas (día 3)

Answer 1

Answer:

1. See below

2. F(x) = -0.10x² + 8x + 200

3. You earn the maximum income of S/320

4. The income decreases

5. x < 0; x > 100

Step-by-step explanation:

1. Strategies

            Let x = the days

Then 100 - x = the kilograms of oranges each day

     and 0.10x = the daily price increase

and 2 + 0.10x = the daily price

Income = (kilograms of oranges) × (price per kilogram)

F(x)= (100 - x)(2 + 0.10x)

2. Another expression for F(x)

You could multiply the factors to get a polynomial in its standard form.

$\begin{array}{rcl}F(x) & = & (100 - x)(2 + 0.10x)\\& = & 200 + 8x - 0.10x^{2}\\& = &\mathbf{ -0.10x^{2} + 8x + 200}\\\end{array}$

3. If sales are made in 20 da

$\begin{array}{rcl}F(x) & = & (100 - x)(2 + 0.10x)\\& = & (100 - 20)(2 + 0.10\times20)\\& = & 80(2 + 2.0)\\& = &80 \times 4\\&=&\textbf{S/320}\\\end{array}\\\text{Sales made on Day 20 will earn the maximum income of } \textbf{S/320}$

4. If sales are made after 40 da

(a) If sales are made on Day 40

$\begin{array}{rcl}F(x) & = & (100 - x)(2 + 0.10x)\\F(x) & = & (100 - 40)(2 + 0.10\times40)\\& = & 60(2 + 4.0)\\& = &60 \times 6\\&=&\textbf{S/360}\\\end{array}$

(b) If sales are made on Day 41

$\begin{array}{rcl}F(x) & = & (100 - 41)(2 + 0.10\times41)\\& = & 59(2 + 4.1)\\& = &59 \times 6.1\\&=&\textbf{S/359.90}\\\end{array}\\\text{The \textbf{income decreases} after Day 40.}$

5. Unuseful parts of the graph

(a) x <0

We can't have negative times because we received the oranges for sale only on Day 0.

(b) x > 100

1 kg of oranges is damaged each day.

After 100 da, there are no more oranges to sell.

The parts of the graph below x = 0 and above x = 100 do not contribute to resolution of the problem.