A store sells two brands of laptop sleeves. The store pays $...

A store sells two brands of laptop sleeves. The store pays $\$ 25$ for each brand $A$ sleeve and $\$ 30$ for each brand $B$ sleeve. $A$ consulting firm has estimated the daily demand equations for these two competitive products to be $x=130-4 p+q$ Demand equation for brand $A$ $y=115+2 p-3 q$ Demand equation for brand $B$ where $p$ is the selling price for brand $A$ and $q$ is the selling price for brand $B$. (A) Determine the demands $x$ and $y$ when $p=\$ 40$ and $q=\$ 50 ;$ when $p=\$ 45$ and $q=\$ 55$ (B) How should the store price each brand of sleeve to maximize daily profits? What is the maximum daily profit? [Hint: $C=25 x+30 y, \quad R=p x+q y,$ and $P=R-C .]$

Transcript

00:01 In the first part of given question, we have to determine the values of x and y for the different values of p.

00:13 We have given that cost of each sluice brand a is equal to dollar 25 and cost of each sluise.

00:37 Brand b is equal to dollar 30 this is given for the store sale of brand laptops sluels also we have given that the demand function we have given that the demand function for brand a for brand a is equal to x is equal to 130 minus 4 times b plus q and demand function for brand and b is equals to 115 plus 2 times p minus 3 times q is equal to y then we have the cost function is then the cost function is c is equal to 25 x plus 30 times y which is equal to 25 times 130 minus 4 p plus plus 30 times 115 plus 2 times b minus 3 times q with simplifying these cost function is equal to 6 ,700 minus 40 times p plus 65 times q we will take given function as brand 1 and brand b as a sorry as the equation number 2 and cost function as the equation number 3 now the revenue function.

02:19 Revenue function is given as r is equal to p times x plus q times 1 which is equals to p times p times 130 minus 4 times p plus q plus q times 1 plus q times 1 plus 2 times p minus 3 times q.

02:44 Therefore we have the revenue function r of x is equals to 130 times p plus 1115 times q minus 4p square minus 3 q square plus 3 times p times q take it as equation number one then we know that the profit function p of x is given as revenue function minus cost function so simplifying it we get the profit function if we we substituted the value of r of x and c of x, we get the profit function in 170 times p plus 180 times q, sorry, plus 180 times q minus 4 times p square minus 3 times q square plus 3 times p times q minus 670.

03:39 We will take it as equation number 5.

03:42 Now for the first part we have if p is equal to dollar 40 and q is equal to dollar 50.

03:51 We have to find x and y.

03:54 Then from equation 1, equation 1 cues us x is equal to 130 minus 4 times 40 plus 50.

04:06 Therefore we have x is equal to 20 and from equation 2, give you.

04:14 Use us y is equal to 1115 plus 2 times 40 minus 3 times 50 this gives us y is equal to 45 now again we have given if we have p is equal to dollar 45 and q is equal to dollar 55 then again from equation number one we get x is equal to 130 minus 4 times 45 plus 55 this gives us x is equal to 5 again from equation number 2 we have y is equal to 150 plus 2 times 45 minus 3 times 55 therefore we have y is equal to 40 which is a required values.

05:14 Now in the next part of the given question, we have to find the daily maximum profit.

05:26 For this, we need to, we need to maximize the equation...

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