Question 1130303

 The price p and the quantity x sold of a certain product obey the demand equation below.

{{{x= -4p+88}}},  {{{0<= p<= 22}}} 
{{{p(x)=(88-x)/4}}}
{{{p(x)=22-x/4}}}

&#8203;(a)
Express the revenue R as a function of x.

{{{R(x)=p(x)*x }}}
{{{R(x)= (22-x/4)x}}}
{{{R(x)= 22x-(1/4)x^2}}}


&#8203;(b)

What is the revenue if 68 units are&#8203; sold?

{{{R(68)= 22*68-(1/4)68^2}}}
{{{R(68)= 1496-1156}}}
{{{R(68)= 340}}}


&#8203;(c)

What quantity x maximizes&#8203; revenue? What is the maximum&#8203; revenue?

{{{R(x)= 22x-(1/4)x^2}}}

{{{R(x)}}} is a quadratic with {{{a=-1/4}}}, {{{b=22}}}
max occurs at {{{x=-b/2a = (-22/(-2/4))= (-22/(-1/2))=22*2=44}}}
max occurs at {{{x=44}}}

&#8203;(d)
What price should the company charge to maximize&#8203; revenue?

{{{p(x)=22-x/4}}}...max occurs at {{{x=44}}}
{{{p(44)=22-44/4}}}
{{{p(44)=22-11}}}
{{{p(44)=11}}}
the company should charge {{{11}}} to maximize&#8203; revenue

&#8203;(e)

What price should the company charge to earn at least &#8203;$288 in&#8203; revenue?

{{{288= p*(-4p+88)}}} 
{{{288= -4p^2+88p }}}
{{{72= -p^2+22p }}}
{{{p^2-22p+72=0}}}
{{{p^2-4p-18p+72=0}}}
{{{(p^2-4p)-(18p-72)=0}}}
{{{p(p-4)-18(p-4)=0}}}
{{{(p-18)(p-4)=0}}}

solutions:
{{{p=4}}} or
{{{p=18}}}