A two and a half pound of grass seeds covers one eight of an acre.
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The objective function is simply a function that is meant to be maximized. Because this function is multivariable, we know that with the applied constraints, the value that maximizes this function must be on the boundary of the domain described by these constraints. If you view the attached image, the grey section highlighted section is the area on the domain of the function which meets all defined constraints. (It is all of the inequalities plotted over one another). Your job would thus be to determine which value on the boundary maximizes the value of the objective function. In this case, since any contribution from y reduces the value of the objective function, you will want to make this value as low as possible, and make x as high as possible. Within the boundaries of the constraints, this thus maximizes the function at x = 5, y = 0.
We start with the more complicated side which is the left side, and show that, on using some trigonometric identities, we will get the term on the right side .
Using Quotient identity for tangent function, we will get
Taking out sine function from the numerator
Cancelling the common term of numerator and denominator