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:
Step-by-step explanation:
We are given that linear differential equation
Auxillary equation
C.F=
P.I=
P.I=
P.I=
and 
where D square is replace by - a square
P.I=
Hence, the general solution
G.S=C.F+P.I
Answer:
33.75
Step-by-step explanation:
12.3 -(-21.45)
12.3 + 21.45 (<em>because</em><em> </em><em>minus </em><em>of </em><em>minus </em><em>becomes</em><em> </em><em>plus)</em>
33.75
He will get 5 pieces of rope, with 2 meters of rope left over. This is because 37 divided by 7 is 5 with a remainder of 2
