Answer:
$21,623.70
Step-by-step explanation:
A suitable financial calculator can compute the beginning balance and the remaining balance for you. The attachments show a TI-Nspire calculator's TVM solver app being used to answer this question.
The first attachment shows the computation of the loan value. It is about $37,624.54.
The second attachment shows the computation of the remaining balance after 16 of the 32 semi-annual payments have been made.
The loan balance 8 years from the end of the loan will be about $21,623.70.
20.if m <6=72, then m < 7=108
21. if m<8,then m<7=80
22.if m<110,then m<6=90
23.if m<123, then m<8=123
24.if m<142, then m<7=38
25.if m<13, then m<8=167
26.if m<170, then m<10
27.if m<26, then m <154
,
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).
Answer:
Step-by-step explanation:
From the question we are told that:
Function given
Co-ordinates
(x,y)=[1, 4]
Generally the second differentiation of function is mathematically given by
Therefore critical point
Generally the substitutions of co-ordinate into function is mathematically given by
For 1
For 4
For critical point 3
Therefore the maximum value of f(x) = –x2 + 6x over the interval [1, 4] is given by
