A farmer owns a paddock that is bounded by part of a river and a straight road running east-west as in the diagram below. The part of the river forming the paddock can be regarded as cubic $(ax^3+bx^2+cx+d)$ in shape. The farmer wishes to divide the paddock into two paddocks of equal area by running a straight fence from point $A$ (the western junction of the river and the road) to point $B$ on the river.
The questions are
a) Dealing with a cubic can be difficult. Approximate this field to a parabola, find the area of the two paddocks so formed and the length of the fence from $A$ to $B$.
b) Using a real field (as a cubic) repeat this, find the area of the two paddocks so formed and the length of the fence from $A$ to $B$.
This is math homework. I'm really stuck and have no idea what to do, please help me; thanks.
Here is a solution for part (b).
We choose a coordinate system with origin $A(0,0)$ at the lower left corner of the picture. Let $$f(x)=ax^3+bx^2+cx +d.$$ Then $$f'(x) = 3ax^2 + 2bx + c. $$
From the given picture we immediately see that $d=0$ because $d$ is the $y$-intercept of the blue cubic curve (the "river").
We can also find $a$, $b$, $c$ from the picture. We simply observe that the cubic curve must go through the points $(30,120)$, $(90,0)$; and the slope $f'(90)$ is zero. This gives us the equations $$ f(30) = a\cdot30^3 + b\cdot 30^2 + c\cdot 30 = 120 \\ f(90) = a\cdot90^3 + b\cdot 90^2 + c\cdot 90 = 0 \\ f'(90) = 3a\cdot90^2 + 2b\cdot90 + c= 0. $$ Solving these equations we find $a = {1\over900}, \ b = -{1\over5}, \ c = 9. \ $ So our cubic curve is $$ f(x)={1\over900}x^3 -{1\over5}x^2 +9x. $$
The total area of the two paddocks (bounded by the river and the road) is $$ \int_0^{90} f(x)\,dx = \int_0^{90} \left({1\over900}x^3 -{1\over5}x^2 +9x \right) dx=6075. $$ Then the area of each of the two new paddocks is $6075/2 = 3037.5$.
We do not know the exact location of point $B$; we will now find it from the condition that the fence $AB$ cuts the field in half. The fence is a straight line that goes through $A(0,0)$, therefore the equation of the fence line is $$ y= kx, $$ where $k$ is unknown. The $x$-coordinate of $B(x_B,f(x_B))$ is connected with $k$ by the equation $$ f(x) = kx, $$ $$ {1\over900}x^3 -{1\over5}x^2 +9x = kx. $$ Solving this equation we find $x=0$ or $x=90\pm30\sqrt{k}$. From the given picture we know that $x_B$ must be positive and less than $90$, hence $$ x_B = 90 - 30\sqrt{k}. $$
The area of the "upper-left" paddock bounded by the river and the fence is $$ \int_0^{x_B} (f(x)-kx) dx = 3037.5, $$ $$ {1\over3600}x_B^4 - {1\over15}x_B^3 + {9-k\over2}x_B^2 = 3037.5. $$ Solving this together with $x_B = 90 - 30\sqrt{k}$, we find $$ k\approx 1.33907, \quad x_B = 90 - 30\sqrt{k} \approx 55.2845, \quad y_B = k \cdot x_B \approx 74.0299, $$ and the length $L$ of the fence $AB$ is $$ L=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2} \approx 92.4\,{\rm m}. $$
... and here is part (a)
Approximating parabola can be chosen in many different ways. We'll use the following equation for our parabola: $$ g(x) = px^2 + qx + r. $$ Let's choose a parabola that passes through the points $A(0,0)$, $C(30,120)$, and $D(60,60)$. Then $r=0$, while for $p$ and $q$ we have the equations $$ p \cdot 30^2 + q \cdot 30 = 120, \\ p \cdot 60^2 + q \cdot 60 = 60. $$ Solving these equations we find $p = -{1\over10}, \ q = 7.$ So our parabola is $$ g(x) = -{1\over10}x^2 + 7x. $$ Unlike the real (cubic) river, our parabola intersects the road ($x$-axis) at the points $(0,0)$ and $(70,0)$ - and not at $(90,0)$.
The approximate total area of two paddocks (the area bounded by the parabola and the road) is $$ \int_0^{70} g(x)\,dx = {17150\over3} \approx 5716.67. $$ The fence line equation is $y=kx$, and the $x$ coordinate of $B$ is connected to $k$ by $$ g(x)=kx, \qquad - {1\over10} x^2 + 7x = kx; $$ solving this equation we find $$ x_B = 70-10k. $$
The area of the "upper-left" paddock bounded by the parabola and the fence is then one-half of the whole field, i.e. ${17150\over6} = 2858\textstyle{1\over3},$ so we have this equation for the area of "upper-left" paddock: $$ \int_0^{x_B} (g(x)-kx) dx = {17150\over6}, $$ $$ -{1\over30}x_B^3 + {7-k\over2} x_B^2 = {17150\over6}. $$ Solving this together with $x_B = 70-10k$, we find $$ k \approx 1.4441, \quad x_B \approx 55.559, \quad y_B = k\cdot x_B \approx 80.233, $$ and the length $L$ of the fence $AB$ is $$ L=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2} \approx 97.6\,{\rm m}. $$