5 Carla packs 4 pairs of pants, 2 pairs of shorts, and 3 skirts for her vacation.
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Let's say "c" is a constant, hmmmm any constant, for any value whatsoever of "x", "y" is always that constant, for example, say c = 3, thus y = 3, so a table for it will look like
now, if you plot those points, it'd looks like the picture below.
now, if you plot those points, it'd looks like the picture below.
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