Answer:
55.4176955.4176955.4176955.41769
Step-by-step explanation:
Answer:
(a)
h=3x
(b)
(c)
(d)
units^3
Step-by-step explanation:
We are given a regular hexagon pyramid
Since, it is regular hexagon
so, value of edge of all sides must be same
The length of the base edge of a pyramid with a regular hexagon base is represented as x
so, edge of base =x
b=x
Let's assume each blank spaces as a , b , c, d
we will find value for each spaces
(a)
The height of the pyramid is 3 times longer than the base edge
so, height =3*edge of base
height=3x
h=3x
(b)
Since, it is in units^2
so, it is given to find area
we know that
area of equilateral triangle is
h=3x
b=x
now, we can plug values
(c)
we know that
there are six such triangles in the base of hexagon
So,
Area of base of hexagon = 6* (area of triangle)
Area of base of hexagon is
(d)
Volume=(1/3)* (Area of hexagon)*(height of pyramid)
now, we can plug values
Volume is
units^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).
