4+4(7x5)= 7/6+58029+>667482+ªº993”QnQ no seeeeeeeeeeeeeeeeeeeeeeeeee
Eeeeeeeeeeeeeee mucho UwU
QwQ
Chauuuuu
-24e^5/6e = -<span>4e^4
hope it helps</span>
Total height of lumber, H = 10 1/2 feet = 21/2 feet .
Height of side panel, h = 5 2/3 feet = 17/3 feet .
Now,
Extra lumber required, L = 2 × Height of side panel - Total height of lumber
![L=[2\times (\dfrac{17}{3})]-\dfrac{21}{2}\\\\L = \dfrac{5}{6}\ feet](https://tex.z-dn.net/?f=L%3D%5B2%5Ctimes%20%28%5Cdfrac%7B17%7D%7B3%7D%29%5D-%5Cdfrac%7B21%7D%7B2%7D%5C%5C%5C%5CL%20%3D%20%5Cdfrac%7B5%7D%7B6%7D%5C%20feet)
Therefore, extra lumber required is 
feet.
Hence, this is the required solution.
Answer: A. A=(1000-2w)*w B. 250 feet
C. 125 000 square feet
Step-by-step explanation:
The area of rectangular is A=l*w (1)
From another hand the length of the fence is 2*w+l=1000 (2)
L is not multiplied by 2, because the opposite side of the l is the barn,- we don't need in fence on that side.
Express l from (2):
l=1000-2w
Substitude l in (1) by 1000-2w
A=(1000-2w)*w (3) ( Part A. is done !)
Part B.
To find the width w (Wmax) that corresponds to max of area A we have to dind the roots of equation (1000-2w)w=0 ( we get it from (3))
w1=0 1000-2*w2=0
w2=500
Wmax= (w1+w2)/2=(0+500)/2=250 feet
The width that maximize area A is Wmax=250 feet
Part C. Using (3) and the value of Wmax=250 we can write the following:
A(Wmax)=250*(1000-2*250)=250*500=125 000 square feets
The 4th one, since both values have less than .5 after the decimal
