Respuesta :
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.
What is profit?
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.
To learn more about profit click here
https://brainly.com/question/21297845
#SPJ2