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ikadub [295]
4 years ago
5

Need help quickly (:

Mathematics
2 answers:
FromTheMoon [43]4 years ago
7 0
30-(increase/0.5)*2=sold
i*s=revenue
i*30-(i^2/0.5)*2=r
I=7.5=>r=0
i=7=>r=14
7/0.5=15 number of increases in the cost
sweet-ann [11.9K]4 years ago
5 0

Answer:

15 is the answer.

Step-by-step explanation:

Initial number of selling = 30 at the cost $2.50 each

on increasing $0.50  cost on each  we have 2 fewer selling

that is number of selling be x so its 30-2x

cost of selling for  30-2x will be = 2.50+0.50x

revenue = (30-2x)(2.5+0.5)

the value of x for which revenue is increased

(30-2x)(2.5+0.5x)>0

(30-2x)(2.50+0.50x)\geq 0\\75+15x-5x-x^2\geq 0\\75+10x-x^2\geq 0\\x^2-10x-75\leq 0\\(x+5)(x-15)\leq 0\\-5\leq x\leq 15\\

so the maximum value of x is 15

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Now, lets put the numbers on the other side.

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