Question

a farmer plans to enclose a rectangular pasture adjacent to a river (see figure). the pasture must contain 45,000 square meters in order to provide enough grass for the herd. no fencing is needed along the river. what dimensions (in m) will require the least amount of fencing? (enter your answers as a comma - separated list.)

Explanation:

Step1: Define variables and equations

Let the length parallel to the river be \(x\) and the width perpendicular to the river be \(y\). The area \(A = xy=45000\), so \(x=\frac{45000}{y}\). The fencing length \(L = x + 2y\).

Step2: Substitute \(x\) into the fencing - length formula

Substitute \(x=\frac{45000}{y}\) into \(L\): \(L(y)=\frac{45000}{y}+2y\), \(y>0\).

Step3: Find the derivative of \(L(y)\)

Using the power rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) (\(u = 45000\), \(v = y\)) and \((ay^{n})^\prime=any^{n - 1}\) (\(a = 2\), \(n = 1\)), \(L^\prime(y)=-\frac{45000}{y^{2}}+2\).

Step4: Set the derivative equal to zero and solve for \(y\)

Set \(L^\prime(y)=0\): \(-\frac{45000}{y^{2}}+2 = 0\). Then \(\frac{45000}{y^{2}}=2\), \(y^{2}=22500\), \(y = 150\) (since \(y>0\)).

Step5: Find \(x\)

Substitute \(y = 150\) into \(x=\frac{45000}{y}\), \(x=\frac{45000}{150}=300\).

Step6: Check the second - derivative (to confirm it's a minimum)

Find the second - derivative \(L^{\prime\prime}(y)=\frac{90000}{y^{3}}\). When \(y = 150\), \(L^{\prime\prime}(150)=\frac{90000}{150^{3}}>0\), so \(L(y)\) has a minimum at \(y = 150\).

Answer:

\(300,150\)