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
The answer to number 23 is D, 49.
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
The decimal is about 1.1 and you can’t simplify it
