Find the dimensions of a rectangle with area 216 m 2 whose perimeter is as small

Question

Find the dimensions of a rectangle with area 216 m2 whose perimeter is as small as possible. (If both values are the same number, enter it into both

The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function

P =

where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?thousand foot-candles

A farmer wants to fence an area of 6 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence?

Explanation / Answer

1)a rectangular of given area which will have minimum perimeter is a square

a=(side)^2

side=sqrt (216)

   =14.696 m

so both values are =14.696 m (ans)

2=

(100 I) / (I^2 + I + 9)
dp/dl = 100[(I^2 + I + 9) - (2l+1)*l]/ (I^2 + I + 9)^2 = 0 >> maxima
[- I^2 + 9] = 0
l = + 3 or l = -3
========================
dp/dl = 100[9 - l^2]/ (I^2 + I + 9)^2
d^2p/dl^2 = 2nd derivative
= 100 { - 2l (I^2 + I + 9)^2 - 2(I^2 + I + 9)(2l+1)(9-l^2)]/(I^2 + I + 9)^4
= 100 { - 2l (I^2 + I + 9) - 2(2l+1)(9-l^2)]/(I^2 + I + 9)^3
= 100[A - B]/C
=====================
for l = 3 > A = - 126, B =0, C = 9261 > d^2p/dl^2 = - 126/9261 = -ve > maximum >>> select this l = 3

for l = - 1 > A = 2, B =0, C = 1 > d^2p/dl^2 = 200 = +ve > minimum---- reject l = - 1
====================
for P(max) value > put l +3 in original equ
P(max) = (100*3) / (9+ 1 + 9) = 300/19 = 15.789 (ans)

3)

= width of the field
q = length

pq = 6 x 10^6 ft

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