# A diamond dealer finds that the demand for flawless diamonds is governed by `p=4x^2-60x+20` , where x is in carats (in units of hundred) with `0ltxlt=5` and p is in thousands of dollars.  Find the number of carats to be made available to maximize revenue. The number of carats is _____ (in units of hundred). What are the price and the revenue for this amount? The price is ______ (in thousands of dollars) and the revenue is ______ (in thousands of dollars). (For the price and the revenue, you may found the calculations easier using a calculator) Note that revenue is the product of the demand function and the number of units sold. So, our revenue function is:

`R = p*x=(4x^2-60x+20)x`

`R = 4x^3-60x^2+20x`

To determine the number of carats (x) that will maximize the revenue, take the derivative of R.

`R' = 12x^2-120x+20`

Set R' equal to zero.

`0=12x^2-120x+20`

Divide the equation by the GCF of 12,20 and 120, to simplify.

`0/4=(12x^2-120x+20)/4`

`0=3x^2-30x+5`

Use the quadratic formula to solve for x.

`x = (-b+-sqrt(b^2-4ac))/(2a) = (-(-30)+-sqrt((-30)^2-4*3*5))/(2*3) =(30+-sqrt840)/6`

`x = (30+sqrt840)/6=9.83 `        and         `x=(30-sqrt840)/6=0.17`

Since we are given with the interval of x which is 0<x<=5, consider only the value of x that fallsto this interval.

Hence, the number of carats that miximimize the revenue is x=0.17 hundreds.

To solve for the price and revenue at x=0.17, substitute this value to the demand fucntion and revenue function.

So, when x = 0.17, p is:

`p = 4x^2-60x+20= 4(0.17)^2-60(0.17)+20= 9.9156`

And R is:

`R=4x^3-60x^2+20x=4(0.17)^3-60(0.17)^2+20(0.17) = 1.6857`

Hence, at x=0.17, the price is \$9.9156 thousands and the revenue is \$1.6857 thousands.

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