
Answer:
the solution is in the explanation
Explanation:
q = D(p) = 0.25(250 - [tex]p^{2}[/tex])
[tex]d_{q}[/tex]/[tex]d_{p}[/tex] = 0.25 (0-2p) = -0.5p
a. Â Elasticity E(p) = [tex]\left[\begin{array}{ccc}p/q\\:\\d_{q}/d_{p} \end{array}\right][/tex]
   E(p) = [tex]\left[\begin{array}{ccc}p/0.25(270-p^{2}) X (-0.5)P \end{array}\right][/tex]
   E(P) = 2[tex]P^{2}[/tex]/270-[tex]P^{2}[/tex]
b. Â Â E(10) =2(100)/ 270 - 100 = 200/170 = 1.18
c. Â Â E(10) > 1
    so the demand is elastic
    hence unit price be lowered to increase revenue
     yes
d. Â Â Â for maximum revenue: E(p) = 1
                            2[tex]P^{2}[/tex]/270-[tex]P^{2}[/tex] = 1
     cross multiply you have     3[tex]P^{2}[/tex] = 270
                            P = [tex](270/3)^{1/2}[/tex]
                            P = 9.48 dollars