Question

ex a former has 1200 ft of fencing for enclosing a trapezoid field along a river as shown, one of the parallel sides is three times longer than the other, no fencing is needed along the river. find the largest area the farmer can enclose
a = (1/2)h(a + b)

Explanation:

Step1: Define Variables and Fencing Constraint

Let the shorter parallel side be \( x \), so the longer is \( 3x \), and the height (and the length of the non - parallel side along the fence) be \( y \). The total fencing is \( x + 3x+ 2y=1200 \) (since there are two vertical sides of length \( y \) and the two parallel sides \( x \) and \( 3x \)). Simplify: \( 4x + 2y = 1200\), then \( 2x + y=600\), so \( y = 600 - 2x \).

Step2: Express Area in Terms of \( x \)

The area of a trapezoid is \( A=\frac{1}{2}h(a + b) \), where \( h = y \), \( a=x \), \( b = 3x \). Substitute: \( A=\frac{1}{2}y(x + 3x)=\frac{1}{2}y(4x)=2xy \). Now substitute \( y = 600 - 2x \) into the area formula: \( A=2x(600 - 2x)=1200x-4x^{2} \).

Step3: Find the Maximum of the Quadratic Function

The function \( A(x)=- 4x^{2}+1200x \) is a quadratic function with \( a=-4 \), \( b = 1200 \), \( c = 0 \). For a quadratic function \( f(x)=ax^{2}+bx + c \), the vertex (which gives the maximum when \( a<0 \)) occurs at \( x=-\frac{b}{2a} \). So \( x=-\frac{1200}{2\times(-4)}=\frac{1200}{8} = 150 \).

Step4: Calculate the Maximum Area

Substitute \( x = 150 \) into the area formula. First, find \( y=600-2\times150 = 300 \). Then \( A=2\times150\times300=90000 \) square feet. We can also use the vertex formula for the quadratic \( A(x)=-4x^{2}+1200x \). The maximum value of \( A \) is \( A(150)=-4\times(150)^{2}+1200\times150=-4\times22500 + 180000=-90000 + 180000 = 90000 \).

Answer:

The largest area the farmer can enclose is \( 90000\space\text{sq ft} \)