Let
x ----------> the height of the whole poster
<span>y ----------> the </span>width<span> of the whole poster
</span>
We need
to minimize the area A=x*y
we know that
(x-4)*(y-2)=722
(y-2)=722/(x-4)
(y)=[722/(x-4)]+2
so
A(x)=x*y--------->A(x)=x*{[722/(x-4)]+2}
Need to minimize this function over x > 4
find the derivative------> A1 (x)
A1(x)=2*[8x²-8x-1428]/[(x-4)²]
for A1(x)=0
8x²-8x-1428=0
using a graph tool
gives x=13.87 in
(y)=[722/(x-4)]+2
y=[2x+714]/[x-4]-----> y=[2*13.87+714]/[13.87-4]-----> y=75.15 in
the answer is
<span>the dimensions of the poster will be
</span>the height of the whole poster is 13.87 in
the width of the whole poster is 75.15 in
Answer:
A.
{7}
Step-by-step explanation:
hello
sqrt(56-7) =sqrt(49)=7
Answer:
distance covered in 2 hours = 80 km
distance covered in 1 hour = 80÷2 = 40 km
distance covered in 7/2 hours = 40 × 7/2
= 140 km
(3 1/2hours = 7/2 hours)
3 = 379.25
4 = 1587.50
5 = 198.88
6 = 1217.50
7 = 1788.20
8 = 1408
9 = 205
10 = 1516
11 = 2034
12 = 880
13 = 1994.57
14 = 1157.77
15 = 38812.50
remember to round the numbers, and also write random numbers everywhere so it looks like you did it.
The asnwer is 32 :) hope it helps
