$18.75 to $18.60 identify the percent of change as an increase or decrease . then find the percent of change round to the neares
2 answers:
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Let u = 5x+1. Then du = 5·dx, and your integral is
Answer:
For a rectangular solid with:
width = w
length = l
height = h
The volume is equal to:
V = w*l*h
in this case we know that:
width = w.
"length of 7 more than the width"
l = w + 7.
"a height that is equivalent to 15 decreased by 3 times the width."
h = 15 - 3*w
Then the volume will be:
V = w*l*h = w*(w + 7)*(15 - 3w) = (w^2 + 7*w)*(15 - 3*w)
V = ( -3*w^3 + 15*w^2 + 105*w - 21*w^2)
V = (-3*w^3 - 6*w^2 + 105*w)
Now, the maximum volume will be for the value of w such that:
V'(w) = 0.
and:
V''(W) < 0
Where:
dV/dw = V'(w).
dV'/dw = V''(w)
Then first we need to differentiate the equation for the volume.
V'(w) = dV/dw = ( 3*(-3*w^2) + 2*(-6*w) + 105)
V'(w) = -9*w^2 - 12*w + 105.
Then we need to find the solution for:
-9*w^2 - 12*w + 105 = 0.
We can use the Bhaskara formula, and we will get:
Then the two solutions are:
w₁ = (+12 - 62.6)/(-18) = 2.81
w₂ = (+12 + 62.6)/(-18) = -15.5
But we can not have a negative width, so we can just discard the second solution.
Now let's check the second condition for the maximum, we must have:
V''(2.81) < 0.
V'' = dV'/dw = 2*(-9*w) - 12 = -18*w - 12
V''(2.81) = -18*2.81 - 12 = -62.58 < 0 .
Then the volume is maximized when w = 2.81, and the maximum volume will be:
V(2.81) = (-3*(2.81)^3 - 6*(2.81)^2 + 105*2.81) = 180.1