question 3. a company runs food service concessions for sporting events throughout the country. their…

question 3. a company runs food service concessions for sporting events throughout the country. their marketing research department chose a particular football stadium to test market a new jumbo hot dog. it was found that the demand for the new hot dog is given approximately by\n\n$p = 4-ln(x),1leq xleq500$,\n\nwhere x is the number of hot dogs (in thousands) that can be sold during one game at a price of p dollars. if the company pays 1 dollar for each hot dog, how should the hot dogs be priced to maximize the profit per game?\n\nquestion 4. a company determines that in order to sell x items, the price per item, in dollars, must be $p(x)=1410$. the company also determines that the total cost, in dollars, to produce x items is given by $c(x)=6400 + 470x+1.8x^{2}$. how many items must the company produce and sell in order to maximize profit?

question 3. a company runs food service concessions for sporting events throughout the country. their marketing research department chose a particular football stadium to test market a new jumbo hot dog. it was found that the demand for the new hot dog is given approximately by\n\n$p = 4-ln(x),1leq xleq500$,\n\nwhere x is the number of hot dogs (in thousands) that can be sold during one game at a price of p dollars. if the company pays 1 dollar for each hot dog, how should the hot dogs be priced to maximize the profit per game?\n\nquestion 4. a company determines that in order to sell x items, the price per item, in dollars, must be $p(x)=1410$. the company also determines that the total cost, in dollars, to produce x items is given by $c(x)=6400 + 470x+1.8x^{2}$. how many items must the company produce and sell in order to maximize profit?

Answer

Explanation:

Step1: Define profit function

Profit ( P(x)=R(x)-C(x) ), where revenue ( R(x)=p(x)\cdot x ). Given ( p(x) = 1410 ), then ( R(x)=1410x ). Cost function ( C(x)=6400 + 470x+1.8x^{2} ). So ( P(x)=1410x-(6400 + 470x+1.8x^{2})= - 1.8x^{2}+940x - 6400 ).

Step2: Find the derivative of profit function

Differentiate ( P(x) ) with respect to ( x ). Using the power rule ( (ax^{n})^\prime=anx^{n - 1} ), ( P^\prime(x)=-3.6x + 940 ).

Step3: Set derivative equal to zero

Set ( P^\prime(x)=0 ), so ( -3.6x + 940=0 ). Solving for ( x ), we get ( 3.6x=940 ), then ( x=\frac{940}{3.6}=\frac{9400}{36}=\frac{2350}{9}\approx261.11 ).

Step4: Check the second - derivative

Differentiate ( P^\prime(x) ) to get ( P^{\prime\prime}(x)=-3.6<0 ). Since the second - derivative is negative, the function ( P(x) ) has a maximum at ( x=\frac{2350}{9}\approx261.11 ). But since ( x ) represents the number of items, we consider the whole number. We can also check the values of ( P(x) ) at ( x = 261 ) and ( x = 262 ).

( P(261)=-1.8\times(261)^{2}+940\times261-6400=-1.8\times68121 + 245340-6400=-122617.8+245340 - 6400=116322.2 )

( P(262)=-1.8\times(262)^{2}+940\times262-6400=-1.8\times68644+246280-6400=-123559.2+246280-6400 = 116320.8 )

Answer:

The company should produce and sell ( 261 ) items to maximize profit.