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# Foundations of finance - Math Problem Example

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The above table shows how the perimeter of the rectangular lawn’s perimeter is computed. Six alternatives were chosen. All alternatives generated to a 200 sq. meter rectangle area (Adler, 2012). Of the different…

## Extract of sample "Foundations of finance"

January 30 Finance Foundations Question1 Table Rectangle Computation The above table shows how to maximize the rectangular courtyard. The above table shows how the perimeter of the rectangular lawn’s perimeter is computed. Six alternatives were chosen. All alternatives generated to a 200 sq. meter rectangle area (Adler, 2012). Of the different alternatives, Alternative B, 100 meters length by 2 meters wide, generates the largest rectangular lawn perimeter, 198.5 meters, contributing to the largest surface area. All the other alternatives generated lesser rectangular lawn perimeters. Alternative B is the best solution. The perimeter formula is:
Perimeter = L +L +W W. The formula for Area is: L x W.
The above table shows the steps used to generate the best alternative. Six 200 sq. meter area alternatives were chosen. Next the required “borders” (2 meters, 1.5 meters, 1 meter) were respectively deducted from each side of the alternatives. Next, the perimeter of the lawn perimeter was generated. The alternative with the biggest perimeter is the correct answer. As expected, alternative B generated the biggest lawn perimeter fence. Alternative B shows the correct length, “L”, should be 100 meters. Likewise, alternative B shows the correct width,” W”, should be 2 meters.
Question 2
Table 2
Computation
Table 3
Computation
The above tables 2 and 3 show the computations for the cable lengths needed for both the 6 meter tower and the 15 meter tower. To reduce total cost, the above table 3 shows that as the junction box should be placed in the middle of the two towers (ten meters width for each of the two towers), total cost = £ 15.60 million. When the junction box is moved 1 meter to the left (Alternative C), the total cost increases to £19.36 million. When the junction box is moved 1 meter to the right (Alternative B), the total cost increases to £19.38 million. Alternative E and F shows as width of either tower is pegged at 1 meter, the total cost reaches more than £ 21.00 million. Based on the above table on the total cost of the different alternatives, alternative A generated the minimum (lowest) total cost, £16.50 million. To reiterate, the junction box should be strategically placed in the middle of the two towers in order to generate the minimum (lowest) total cost.
The steps taken arrive at the above answer is shown in the excel file. The math formula, C2 = A2 + B2, was used to arrive at the required cable length of each tower (Wheater, 2012). The “C” variable represents the cable length of each cable. The “A” variable represents each tower’s height. The “B” variable represents each tower’s width. After getting the cable length, the cost of each tower’s cable length is computed. The cost of each tower is multiplied by the cable purchased to arrive at the total cable cost of both towers. Consequently, the table 2 shows the total cost of both towers. The cost of each cable is multiplied by the cable length to arrive at the total cable cost of both towers.
References:
Adler, D.,2012, Perimeter, Area, and Volume. New York: Holiday House Press.
Wheater, C.,2012, Practice Makes Perfect Basic Math. London: McGraw-Hill Press. Read More
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