Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Juan is debating whether or not to continue the recent lunch special promotion at his restaurant. The following graph shows the profit earned, y, from the number of lunch specials ordered per hour, x.


Use the graph to complete the following statements. If necessary, round to the nearest whole number.


The maximum profit earned from lunch special orders per hour is about $_____.
The restaurant will break even if ______ lunch specials are ordered.

Answer:

It is given that y represents the profit earned, and x represents the number of lunch specials ordered at Juan's restaurant per hour.

To determine the maximum profit Juan's restaurant will earn from lunch specials, estimate the highest y-value from the graph. From the graph, it can be seen that the maximum y-value is halfway between 110 and 120. So, the maximum profit Juan's restaurant will earn from lunch specials per hour is about $115.

Next, to determine when the restaurant will break even, observe the x-value when y = 0. It can be seen that graph crosses the x-axis in two places, when x = -5 and x = 10. Since x represents the number of lunch specials ordered, the x-value must be positive. So, the restaurant will break even after 10 lunch specials are ordered.

Lastly, to determine the interval on which the restaurant makes a profit, observe from the graph where the values of x and y are positive. So, the interval on which the restaurant makes a profit is (0, 10).

Step-by-step explanation:

115, 10, (0,10)

^those are the answers

The maximum profit earned from lunch special orders per hour is about [tex]\$ 112.5[/tex]. The restaurant will break even if [tex]10[/tex] lunch specials are ordered.

Profit is the amount which we earned after selling something more that its cost price.

We have,

Profit earned [tex]=y[/tex],

The number of lunch specials ordered per hour [tex]=x[/tex]

Now,

From the graph,

Profit(y)  [tex]=a(x+5)(x-10)[/tex]

[tex]y=a[x^2-10x+5x-50][/tex]

[tex]y=a[x^2-5x-50][/tex]

Now,

When [tex]x=0[/tex]  ,

then,

[tex]y=-50a[/tex]     [tex].....(i)[/tex]

And,

From graph When [tex]x=0[/tex] ,

[tex]y=100[/tex]     [tex].....(ii)[/tex]

So, from above equations ,

[tex]-50a=100[/tex]

[tex]a=-2[/tex],

Now,

[tex]y=a[x^2-5x-50][/tex]

Substituting value of a,

[tex]y=[-2x^2+10x+100][/tex]

[tex]y=[-2(x^2-5x)+100][/tex]

[tex]y=[-2(x^2-5x-(\frac{5}{2})^2 )+100+2(\frac{5}{2})^2][/tex]

[tex]y=[-2(x-\frac{5}{2} )^2+112\frac{1}{2}[/tex]

So,

Now

The maximum profit will be ,

From graph,

When [tex]x=\frac{5}{2} = 2.5[/tex]

So,

Putting value of x,

We get,

[tex]y=112.5[/tex]

So, the maximum profit earned from lunch special orders per hour is about [tex]\$ 112.5[/tex].

Now,

When [tex]y=0[/tex] then [tex]x=-5[/tex] or [tex]x=10[/tex]

As [tex]x > 0[/tex]

So,

The restaurant will break even if [tex]10[/tex] lunch specials are ordered.

Hence we can say that the maximum profit earned from lunch special orders per hour is about [tex]\$ 112.5[/tex]. The restaurant will break even if [tex]10[/tex] lunch specials are ordered.

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