Question

1. the profit that a company will make by selling t - shirts will depend on the price they sell the shirt at. price the shirt too high, and not many people will buy. price the shirt to low and lots of people will buy, but you will not earn as much money overall. the relationship between profit and price can be given as: (4 marks)

y = -30.12x² + 1234.92x - 657.93
where y represents the profit and and x represents the selling price of each t - shirt.
a) what is the maximum profit that the company can earn?
b) at what price should they each t - shirt for to achieve this profit?
c) what would the profit of the company be if they priced each shirt at $28?
d) between which two prices should the company price the shirt so that they earn a profit of at least $8000?

Explanation:

Part (a)

Step 1: Identify the vertex of the parabola

The profit function is a quadratic function in the form \( y = ax^2 + bx + c \), where \( a = -30.12 \), \( b = 1234.92 \), and \( c = -657.93 \). For a quadratic function \( y = ax^2 + bx + c \) with \( a < 0 \), the vertex represents the maximum point. The x - coordinate of the vertex of a parabola given by \( y=ax^{2}+bx + c\) is \( x=-\frac{b}{2a} \), and the y - coordinate (which is the maximum profit in this case) can be found by substituting this x - value back into the function.

First, find the x - coordinate of the vertex:
\( x=-\frac{b}{2a}=-\frac{1234.92}{2\times(-30.12)} \)
\( x = \frac{1234.92}{60.24}\approx20.5 \)

Step 2: Find the maximum profit

Now substitute \( x\approx20.5 \) into the profit function \( y=-30.12x^{2}+1234.92x - 657.93 \)

\( y=-30.12\times(20.5)^{2}+1234.92\times20.5-657.93 \)

First, calculate \( (20.5)^{2}=420.25 \)

\( -30.12\times420.25=-30.12\times420 + (-30.12)\times0.25=-12648-7.53=-12655.53 \)

\( 1234.92\times20.5 = 1234.92\times(20 + 0.5)=1234.92\times20+1234.92\times0.5 = 24698.4+617.46 = 25315.86 \)

Now, \( y=-12655.53 + 25315.86-657.93 \)

\( y=(25315.86-12655.53)-657.93=12660.33 - 657.93=12002.4 \)

Part (b)

Step 1: Recall the formula for the x - coordinate of the vertex

For a quadratic function \( y = ax^{2}+bx + c \), the x - coordinate of the vertex (which gives the price that maximizes profit) is \( x =-\frac{b}{2a} \)

We have \( a=-30.12 \) and \( b = 1234.92 \)

\( x=-\frac{1234.92}{2\times(-30.12)}=\frac{1234.92}{60.24}\approx20.5 \)

Part (c)

Step 1: Substitute \( x = 28 \) into the profit function

We have \( y=-30.12x^{2}+1234.92x - 657.93 \), substitute \( x = 28 \)

\( y=-30.12\times(28)^{2}+1234.92\times28-657.93 \)

First, calculate \( 28^{2}=784 \)

\( -30.12\times784=-30.12\times700+(-30.12)\times84=-21084-2530.08=-23614.08 \)

\( 1234.92\times28=1234.92\times(20 + 8)=1234.92\times20+1234.92\times8 = 24698.4+9879.36 = 34577.76 \)

Now, \( y=-23614.08+34577.76 - 657.93 \)

\( y=(34577.76-23614.08)-657.93=10963.68-657.93 = 10305.75 \)

Part (d)

Step 1: Set up the inequality

We want to find the values of \( x \) for which \( y\geq8000 \), so we set up the inequality:

\( -30.12x^{2}+1234.92x - 657.93\geq8000 \)

Subtract 8000 from both sides:

\( -30.12x^{2}+1234.92x-8657.93\geq0 \)

Multiply both sides by - 1 (and reverse the inequality sign):

\( 30.12x^{2}-1234.92x + 8657.93\leq0 \)

Step 2: Solve the quadratic equation \( 30.12x^{2}-1234.92x + 8657.93 = 0 \)

Use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 30.12 \), \( b=-1234.92 \), \( c = 8657.93 \)

First, calculate the discriminant \( D=b^{2}-4ac \)

\( D=(-1234.92)^{2}-4\times30.12\times8657.93 \)

\( (-1234.92)^{2}=1234.92\times1234.92\approx1525027.4064 \)

\( 4\times30.12\times8657.93 = 120.48\times8657.93\approx1043000 \)

\( D\approx1525027.4064 - 1043000=482027.4064 \)

\( \sqrt{D}\approx\sqrt{482027.4064}\approx694.28 \)

Now, \( x=\frac{1234.92\pm694.28}{2\times30.12} \)

First, for the plus sign:

\( x_1=\frac{1234.92 + 694.28}{60.24}=\frac{1929.2}{60.24}\approx32.02 \)

For the minus sign:

\( x_2=\frac{1234.92-694.28}{60.24}=\frac{540.64}{60.24}\approx8.97 \)

Since the quadratic function \( y = 30.12x^{2}-1234.92x + 8657.93 \) opens upwards (because \( a = 30.12>0 \)), the inequality \( 30.12x^{2}-1234.92x + 8657.93\leq0 \) is satisfied for \( x \) between \( x_2\approx8.97 \) and \( x_1\approx32.02 \)

Answer:

s:
a) The maximum profit is approximately \(\$12002.4\)

b) The price per t - shirt should be approximately \(\$20.5\)

c) The profit when the price is \(\$28\) is \(\$10305.75\)

d) The company should price the shirt between approximately \(\$8.97\) and \(\$32.02\)