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Let n be the number of increments by $0.25.
Then the price for each donut is p(n) = 3 + 0.25n and the number of the sold donutes is 1400-70n.
Hence, the revenue is
R(n) = p(n)*(1400 - 70n) = (3 + 0.25n)*(1400 - 70n).
It is a quadratic function , presented as the product of the two linear binomial.
The quadratic function has the roots there, where the linear binomials are zer0, i.e. at
n = -12 and n= 20.
The maximum of the quadratic function is achieved at the midpoint between its roots, i.e. at = 4.
So, 4 (four) increases at $0.25 each are required to get the maximum revenue.
Thus the optimum price per donut is 3 + 4*0.25 = 4 dollars.
The optimum sale is then 1400 - 4*70 = 1120 donuts, and the maximum revenue is
4*1120 = 4480 dollars.
Compare it with the starting revenue of 3*1400 = 4200 dollars.
Solved.