Question 798189
If you need to show work, just write:
{{{f(x)=x(34 - 0.05x)}}}
{{{f(x)=-0.05x^2+34x)}}}
{{{-0.05<0}}} so maximum at {{{x=(-34)/(2*(-0.05))=(-34)/(-0.1)=34/0.1=highlight(340)}}}
 
UNDERSTANDING THE PROBLEM:
The price per pair of running shoes is
${{{34}}} if {{{x<70}}} or
${{{(34 - 0.05x)}}}  if {{{70<=x<=600}}}
They must be sure their customers will not place large orders, because for an order of 600  pairs, the price per pair would be
${{{(34 -0.05*600)}}}= ${{{34-30}}}= ${{{4}}}
 
Anyway, if the company sells {{{x}}} pairs, between 70 and 600 pairs ({{{70<=x<=600}}}), they get
${{{x(34 - 0.05x)}}}
The amount (in $) they get, as a function of {{{x}}}, is a quadratic function:
{{{f(x)=x(34 - 0.05x)}}} or {{{f(x)=-0.05x^2+34x)}}}
For this problem, that function is only defined for some values of {{{x}}}, {{{70<=x<=600}}}. (The domain of that function is restricted by {{{70<=x<=600}}}).
The function {{{f(x)=x(34 - 0.05x)}}} without restrictions graphs as a parabola,
a nicely symmetrical curve, with a maximum exactly halfway between the zeros,
which are at {{{x=0}}},
and where {{{34-0.05x=0}}} <--> {{{0.05x=34}}} <--> {{{x=34/0.05}}} <--> {{{x=680}}}.
So that maximum is at exactly {{{x=(0+340)/2=highlight(340)}}}.
Easy. No memorized formulas required.
 
WHAT IS EXPECTED:
Your teacher may expect you to multiply to get from
{{{f(x)=x(34 - 0.05x)}}} to {{{f(x)=-0.05x^2+34x)}}},
and then apply memorized facts and formulas.
You've been taught the fact that a quadratic function {{{y=ax^2+bx+c}}}
has a maximum or minimum at {{{x=(-b)/2a}}},
which is a maximum if {{{a<0}}}, and is a minimum if {{{a>0}}}.
In this case {{{a=-0.05}}}, so there is a maximum,
and {{{x=(-b)/2a}}} calculates as {{{x=(-34)/(2(-0.05))=(-34)/(-0.1)=34/0.1=highlight(340)}}}