Question

T h e g y m c h a r g e s $ 1 5 0 f o r a y e a r l y m e m b e r s h i p . T h e r e a r e c u r r e n t l y 5 0 0 m e m b e r s . F o r e v e r y $ 5 i n c r e a s e i n p r i c e , t h e g y m w i l l l o s e 1 0 m e m b e r s . a ) H o w m u c h s h o u l d t h e g y m c h a r g e t o m a x i m i z e i t s r e v e n u e ? b ) W h a t i s t h e m a x i m u m r e v e n u e ?

Answer 1

the yearly charges of the gym is 150 $

the number of members are 500

so the total revenue is

500 * 150 = 75000 $

let the x times increase 5 $

then the members lose by gym is 10x

so we can write the equation

(150 + 5x) * (500 - 10x) = 75000 $

solve the LHS of equation

$75000-1500x+2500x-50x^2=0$$50x^2-1000x-75000=0$

it is a quadratic equation,

the maximum value of a quadratic equation is given by ' x = -b/2a

for

$ax^2+bx+c=0$

solve x = -b/2a

$x=\frac{-1000}{2\times50}=10$

so the gym charges should be

150+(10 * 5) = 200 $

a) so the new gys charges is 200 $.

and the number of member will be

500 - (10 * 10 ) = 500 - 100 = 400

b) and the maximum revenue will be

400 * 200 = 80000 $.