A closed circle means that value is included, and an open circle means that value is excluded.
Example:
x > 2 ---- Graphed with an open circle, everything greater than but not including 2
x ≤ 3 ------ Graphed with a closed circle, everything less than and including 3
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
<span> first, write the equation of the parabola in the required form: </span>
<span>(y - k) = a·(x - h)² </span>
<span>Here, (h, k) is given as (-1, -16). </span>
<span>So you have: </span>
<span>(y + 16) = a · (x + 1)² </span>
<span>Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15): </span>
<span>-15 + 16 = a· (0 + 1)² </span>
<span>.·. a = 1 </span>
<span>.·. the equation of the parabola in vertex form is </span>
<span>y + 16 = (x + 1)² </span>
<span>The x-intercepts are the values of x that make y = 0. So, let y = 0: </span>
<span>0 + 16 = (x + 1)² </span>
<span>16 = (x + 1)² </span>
<span>We are trying to solve for x, so take the square root of both sides - but be CAREFUL! </span>
<span>± 4 = x + 1 ...... remember both the positive and negative roots of 16...... </span>
<span>Solving for x: </span>
<span>x = -1 + 4, x = -1 - 4 </span>
<span>x = 3, x = -5. </span>
<span>Or, if you prefer, (3, 0), (-5, 0). </span>
Absolute value= opposite. (positive turns into negative)(negative turns into positive)
