Material=wall+2*side and material is 40 ft so:
40=w+2s
w=40-2s
Area=ws, using w from above we get:
A=(40-2s)s
A=40s-2s^2
dA/ds=40-4s and d2A/ds2=-4
Since d2A/ds2 is a constant negative acceleration, when dA/ds=0, A(s) is at an absolute maximum.
dA/ds=0 when 4s=40, s=10 ft
And since w=40-2s, w=20 ft
So the dimensions of the pen are 20 ft by 10 ft, with the 20 ft side being opposite the wall. And the maximum possible area is thus 200 ft^2
Since the radius is the same but the height changes, the volume would change by the multiple of the height.
In this case the volume of the larger cone would be 4 times as large as the smaller cone so you would need 4 small cones
Answer:

Step-by-step explanation:

here,
a = 1,
b = k
c = 9
now,


Answer:
t=x+44
Step-by-step explanation:
the x ounces are being added to the vase's pre-existing 44 ounces so therefore the equation is t=x+44