Respuesta :
Answer:
a) [tex] 600 = 3v+h[/tex]
The 3 is because we have 3 parallel pieces.
And then if we solve for h we got:
[tex]h = 600 - 3v[/tex]
b) [tex] A = v (600-3v) = 600v - 3v^2[/tex]
And if we analyze the function we see that [tex] v \leq 200[/tex] since we can't have a negative area.
c) [tex] h = 450 , v = 50 [/tex]
And in order to find the maximum area enclosed we need to derivate respect h, the original function.
[tex] \frac{dA}{dv}=600 - 6v =0[/tex]
And as we can see then v= 100 and h = 600-3(100) = 300
And the maximum area would be given by :
[tex] A = hv = 300*100 = 30000[/tex]
Step-by-step explanation:
For this case we can assume the following notation:
[tex] v [/tex] represent the vertical dimension
[tex] h[/tex] represent the horizontal dimension
We know also that the total length of fencing is 600
Part a
So then we can create the following function:
[tex] 600 = 3v+h[/tex]
The 3 is because we have 3 parallel pieces.
And then if we solve for h we got:
[tex]h = 600 - 3v[/tex]
Part b
For this case since we have a rectangular area we have the following formula for the area:
[tex] A = v h[/tex]
[tex] A = v (600-3v) = 600v - 3v^2[/tex]
And if we analyze the function we see that [tex] v \leq 200[/tex] since we can't have a negative area.
Part c
For this case we want to find the dimensions that maximize a total area of 22500, so we have this expression:
[tex] 22500 = 600 v - 3v^2 [/tex]
We cna express this function like this:
[tex] 3v^2 -600 v +22500 = 0[/tex]
And as we can see we have a quadratic equation and we can solve it with the following formula:
[tex] x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
For this case [tex] a = 3, b = -600, c = 22500[/tex]
And if we replace we got:
[tex] v = \frac{600 \pm \sqrt{(600)^2 -4(3)(22500)}}{2(3)}[/tex]
[tex] v_1= \frac{600-300}{6}=50[/tex]
[tex] v_2= \frac{600+300}{6}=150[/tex]
If we use 150 for v then the value of h would be 0 and that not makes sense so then the best solution would be v = 50
And then we can solve for h like this:
[tex]h = 600 - 3v= 600 - 3(50) = 450[/tex]
[tex] h = 450 , v = 50 [/tex]
And in order to find the maximum area enclosed we need to derivate respect h , the original function.
[tex] \frac{dA}{dv}=600 - 6v =0[/tex]
And as we can see then v= 100 and h = 600-3(100) = 300
And the maximum area would be given by :
[tex] A = hv = 300*100 = 30000[/tex]
