Answer:
we have
y=16-4x²
4x²=16
x²=
x=
x=±2
PLEASE HELP ASAPPP!!!
2 answers:
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Let
e = Gibson Explorer’s = 20
v = Gibson Flying V’s
So, our problem is
<u>Maximize</u>
Money = 80e + 5v<span>
<u>Subject to</u>
0 </span>≤ <span>e ≤ 20
</span>0 ≤ v ≤ 20
0 ≤ e + v ≤ 30
In order to solve this problem, we look at the graph (attached), and find the value of Money =80e + 5v at corner points to find the maximum value of money.
(e,v)=(0,0) >> Money = 80e+5v = 80*0+5*0 = 0
(e,v)=(0,20) >> Money = 80e+5v = 80*0+5*20 = 100
(e,v)=(20,0) >> Money = 80e+5v = 80*20+5*0 = 1600
(e,v)=(20,10) >> Money = 80e+5v = 80*20+5*10 = 1650 (maximum)
(e,v)=(10,20) >> Money = 80e+5v = 80*10+5*20 = 900
So Bob can make the <u>most money = $1,650</u> when he makes and sell
e = <span>Gibson Explorer’s = 20
</span>v = Gibson Flying V’s = 10
e = Gibson Explorer’s = 20
v = Gibson Flying V’s
So, our problem is
<u>Maximize</u>
Money = 80e + 5v<span>
<u>Subject to</u>
0 </span>≤ <span>e ≤ 20
</span>0 ≤ v ≤ 20
0 ≤ e + v ≤ 30
In order to solve this problem, we look at the graph (attached), and find the value of Money =80e + 5v at corner points to find the maximum value of money.
(e,v)=(0,0) >> Money = 80e+5v = 80*0+5*0 = 0
(e,v)=(0,20) >> Money = 80e+5v = 80*0+5*20 = 100
(e,v)=(20,0) >> Money = 80e+5v = 80*20+5*0 = 1600
(e,v)=(20,10) >> Money = 80e+5v = 80*20+5*10 = 1650 (maximum)
(e,v)=(10,20) >> Money = 80e+5v = 80*10+5*20 = 900
So Bob can make the <u>most money = $1,650</u> when he makes and sell
e = <span>Gibson Explorer’s = 20
</span>v = Gibson Flying V’s = 10
Check the picture below.
is simply a 9x9 square, so its perimeter is simply 4*9.
