$3820 minimum cost for enclosing the area.
Let's first create a function to express the cost of the area based upon the length of shrubs. We'll use the variable S for the length of the shrubs. So the width will be:
W = 3000/S
The cost of the rectangle will be the cost of fencing multiplied by (2 * W + S) plus the cost of the shrubs multiplied by S. So
C = 16(2*W + S) + 22S
Substitute 3000/S for W giving:
C = 16(2*3000/S + S) + 22S
C = 16(6000/S + S) + 22S
Now simplify
C = 16*(6000/S + S) + 22*S
C = 96000/S + 16S + 22S
C = 96000/S + 38S
Now since you're looking for a maximum, that will only happen whtn the slope of the cost function is 0. And the slope of the function is defined as the first derivative of the function. So let's calculate the first derivative and then solve for 0. For simple functions like the above, what you do is multiply each terms coefficient by the exponent, then subtract 1 from the exponent. So:
C = 96000/S + 38S
Let's make the exponents explicit:
C = 96000S^(-1) + 38S^1
C' = -1*96000S^(-1 -1) + 1*38S^(1-1)
C' = -96000S^(-2) + 38S^(0)
C' = -96000/S^2 + 38*1
C' = -96000/S^2 + 38
Now solve for 0
C' = -96000/S^2 + 38
0 = -96000/S^2 + 38
96000/S^2 = 38
96000 = 38S^2
96000/38 = S^2
48000/19 = S^2
+/- sqrt(48000/19) = S
Since a negative length doesn't make sense for this problem will say that the best length for the shrub line is sqrt(48000/19) or approximately 50.262469 feet. That will make the other dimension 3000/50.262469 = 59.68668193 feet. So the total cost will be
C = 16*(2*59.68668193 + 50.262469) + 22*50.262469
C = 16*(119.3733639 + 50.262469) + 22*50.262469
C = 16*169.6358329 + 22*50.262469
C = 2714.173326 + 1105.774318
C = 3819.947644
And rounding gives 3820.
