[tex]P=-\frac{1}{20}x^2+5x-30[/tex]
a) Use the equation above to complete the table: substitute the x in the equation by the given values in the tabe to find the corresponding value of P:
[tex]\begin{gathered} x=0 \\ P=-\frac{1}{20}(0)^2+5(0)-30 \\ P=-30 \\ \\ x=20 \\ P=-\frac{1}{20}(20)^2+5(20)-30 \\ P=-20+100-30 \\ P=50 \\ \\ x=30 \\ P=-\frac{1}{20}(30)^2+5(30)-30 \\ P=-\frac{900}{20}+150-30 \\ P=-45+150-30 \\ P=75 \end{gathered}[/tex][tex]\begin{gathered} x=50 \\ P=-\frac{1}{20}(50)^2+5(50)-30 \\ P=-\frac{2500}{20}+250-30 \\ P=-125+250-30 \\ P=95 \\ \\ x=60 \\ P=-\frac{1}{20}(60)^2+5(60)-30 \\ P=-\frac{3600}{20}+300-30 \\ P=-180+300-30 \\ P=90 \\ \\ x=90 \\ P=-\frac{1}{20}(90)^2+5(90)-30 \\ P=-\frac{8100}{20}+450-30 \\ P=-405+450-30 \\ P=15 \end{gathered}[/tex]
Table:
b) Graph:
c)
i) maximum possible profit is the maximum P in the vertex: 95
ii) the numer of glasses that need to be sold to make the maximum profit is the x value in the vertex: 50
iii) the number of glasses that need to be sold to make a profit of at least 80; x coordinate when P is 80 in the increasing part of the parabola: Approximately 33
iv) the amount of money initially invested by the three students, value of P1when x is 0: it is negative bcause is athe amount invested -30, They invested 30 Swiss Francs
d)Baljeet : Jane : Fiona
1: 2 : 3
The profit is divided into 6 parts, Baljeet receives 1/6, Jane 2/6 and Finona 3/6
If they sold 40 gasses of lemodnade the profit is 90Swiss Francs
[tex]\frac{90}{6}=15[/tex]
1/6 of 90 is 15, as Fiona received 3/6, she has:
[tex]15*3=45[/tex]
Fiona's share 45 Swiss Francs
