Let the length of the rectangular field along the river be LL (in meters), and the height (width) of the field perpendicular to the river be HH (in meters). Given: The area of the field is A=L×H=540 m2A = L \times H = 540 \, \text{m}^2. The cost of fencing material along the river (the side with length LL) is $25 per meter. The cost of fencing material for the other two sides (each of length HH) is $10 per meter. Objective: We need to find the values of LL and HH that minimize the total cost of the fencing. Step 1: Express the total cost The fencing costs for each side are: One side (length LL) against the river: 25L25L dollars. Two other sides (heights HH): 2×10H=20H2 \times 10H = 20H dollars. So, the total cost CC is: C=25L+20HC = 25L + 20H Step 2: Use the area constraint From the given area of the field, we know that: L×H=540L \times H = 540 This can be rearranged to express HH in terms of LL: H=540LH = \frac{540}{L} Step 3: Substitute HH into the cost function Substitute H=540LH = \frac{540}{L} into the cost function CC: C=25L+20(540L)C = 25L + 20 \left( \frac{540}{L} \right) C=25L+10800LC = 25L + \frac{10800}{L} Step 4: Minimize the cost function To find the value of LL that minimizes the cost, take the derivative of CC with respect to LL and set it equal to zero: dCdL=25−10800L2\frac{dC}{dL} = 25 - \frac{10800}{L^2} Set the derivative equal to zero: 25−10800L2=025 - \frac{10800}{L^2} = 0 Solve for LL: 10800L2=25\frac{10800}{L^2} = 25 L2=1080025=432L^2 = \frac{10800}{25} = 432 L=432≈20.7846 mL = \sqrt{432} \approx 20.7846 \, \text{m} Step 5: Find the corresponding HH Now that we have L≈20.7846L \approx 20.7846, substitute this back into the equation for HH: H=540L=54020.7846≈26 mH = \frac{540}{L} = \frac{540}{20.7846} \approx 26 \, \text{m} Step 6: Verify that this is a minimum To confirm that this value of LL corresponds to a minimum, we check the second derivative of CC: d2CdL2=21600L3\frac{d^2C}{dL^2} = \frac{21600}{L^3} Since L>0L > 0, the second derivative is positive, indicating that the cost function is concave up and L≈20.7846L \approx 20.7846 m gives a minimum. Final Answer: The length of the field along the river is approximately 20.78 meters. The height of the field is approximately 26 meters. These dimensions minimize the cost of fencing.