{"id":23887,"date":"2025-06-17T22:21:45","date_gmt":"2025-06-17T22:21:45","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=23887"},"modified":"2025-06-17T22:21:51","modified_gmt":"2025-06-17T22:21:51","slug":"through-market-research-a-company-finds-that-it-can-expect-to-sell-45-5x-products-if-each-is-priced-at-1-25x-dollars","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/through-market-research-a-company-finds-that-it-can-expect-to-sell-45-5x-products-if-each-is-priced-at-1-25x-dollars\/","title":{"rendered":"Through market research, a company finds that it can expect to sell 45 &#8211; 5x products if each is priced at 1.25x dollars"},"content":{"rendered":"\n<p>Through market research, a company finds that it can expect to sell 45 &#8211; 5x products if each is priced at 1.25x dollars.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s analyze the problem step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Demand function (quantity sold) = 45\u22125&#215;45 &#8211; 5x<\/li>\n\n\n\n<li>Price per product = 1.25&#215;1.25x dollars<\/li>\n<\/ul>\n\n\n\n<p>Here, xx is some variable related to pricing (likely a price factor or a decision variable).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">1. Expression for Revenue<\/h3>\n\n\n\n<p>Revenue RR is given by: R=(Price&nbsp;per&nbsp;product)\u00d7(Number&nbsp;of&nbsp;products&nbsp;sold)R = (\\text{Price per product}) \\times (\\text{Number of products sold})<\/p>\n\n\n\n<p>Substitute the given expressions: R=(1.25x)\u00d7(45\u22125x)=1.25x\u00d7(45\u22125x)R = (1.25x) \\times (45 &#8211; 5x) = 1.25x \\times (45 &#8211; 5x)<\/p>\n\n\n\n<p>Simplify: R=1.25x\u00d745\u22121.25x\u00d75x=56.25x\u22126.25x2R = 1.25x \\times 45 &#8211; 1.25x \\times 5x = 56.25x &#8211; 6.25x^2<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. Interpretation<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>xx affects both the price and the number of units sold.<\/li>\n\n\n\n<li>As xx increases, price per product increases (since price = 1.25&#215;1.25x).<\/li>\n\n\n\n<li>However, the number sold decreases linearly by 5x5x.<\/li>\n\n\n\n<li>The revenue is a quadratic function in xx, opening downward (because the coefficient of x2x^2 is negative: \u22126.25-6.25), which means it has a maximum point.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3. Finding the Price and Quantity that Maximizes Revenue<\/h3>\n\n\n\n<p>To maximize revenue, differentiate R(x)R(x) with respect to xx and set to zero: dRdx=56.25\u221212.5x=0\\frac{dR}{dx} = 56.25 &#8211; 12.5x = 0<\/p>\n\n\n\n<p>Solve for xx: 12.5x=56.25\u2005\u200a\u27f9\u2005\u200ax=56.2512.5=4.512.5x = 56.25 \\implies x = \\frac{56.25}{12.5} = 4.5<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">4. Calculate optimal price and quantity sold<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Price:<\/li>\n<\/ul>\n\n\n\n<p>P=1.25\u00d74.5=5.625&nbsp;dollarsP = 1.25 \\times 4.5 = 5.625 \\text{ dollars}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Quantity:<\/li>\n<\/ul>\n\n\n\n<p>Q=45\u22125\u00d74.5=45\u221222.5=22.5&nbsp;unitsQ = 45 &#8211; 5 \\times 4.5 = 45 &#8211; 22.5 = 22.5 \\text{ units}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">5. Maximum Revenue<\/h3>\n\n\n\n<p>Calculate revenue at x=4.5x = 4.5: R=56.25\u00d74.5\u22126.25\u00d7(4.5)2=253.125\u2212126.5625=126.5625&nbsp;dollarsR = 56.25 \\times 4.5 &#8211; 6.25 \\times (4.5)^2 = 253.125 &#8211; 126.5625 = 126.5625 \\text{ dollars}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Summary<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The company\u2019s revenue as a function of xx is:<\/li>\n<\/ul>\n\n\n\n<p>R=56.25x\u22126.25x2R = 56.25x &#8211; 6.25x^2<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Revenue is maximized when x=4.5x = 4.5.<\/li>\n\n\n\n<li>At this xx, the optimal price per product is <strong>$5.63<\/strong> (rounded), and the expected sales volume is <strong>22.5 units<\/strong>.<\/li>\n\n\n\n<li>The maximum revenue achievable is about <strong>$126.56<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300 words):<\/h3>\n\n\n\n<p>The problem describes a company\u2019s expected sales quantity and pricing strategy through a variable xx. The demand function is 45\u22125&#215;45 &#8211; 5x, indicating that as xx increases, fewer products are sold \u2014 this models the typical negative relationship between price and quantity demanded. The price per product is 1.25&#215;1.25x, which means the price increases proportionally with xx.<\/p>\n\n\n\n<p>To find total revenue, multiply the number of products sold by the price per product, yielding a quadratic expression: R(x)=56.25x\u22126.25x2R(x) = 56.25x &#8211; 6.25x^2. This quadratic opens downward, indicating that revenue increases with xx up to a point and then decreases. To find the maximum revenue, we differentiate R(x)R(x) with respect to xx and set the derivative to zero. Solving this derivative gives the critical value x=4.5x=4.5.<\/p>\n\n\n\n<p>Substituting x=4.5x=4.5 back into the price and quantity functions gives the price of approximately $5.63 and a sales volume of 22.5 units. This combination maximizes revenue, which is about $126.56.<\/p>\n\n\n\n<p>This analysis helps the company understand the optimal pricing strategy to maximize revenue, balancing the trade-off between increasing prices and decreasing sales volume. Pricing too high will decrease quantity sold excessively, reducing revenue; pricing too low limits revenue per unit. This mathematical model guides strategic pricing decisions to optimize income.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Through market research, a company finds that it can expect to sell 45 &#8211; 5x products if each is priced at 1.25x dollars. The correct answer and explanation is: Let&#8217;s analyze the problem step-by-step. Given: Here, xx is some variable related to pricing (likely a price factor or a decision variable). 1. Expression for Revenue [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-23887","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23887","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=23887"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23887\/revisions"}],"predecessor-version":[{"id":23889,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23887\/revisions\/23889"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=23887"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=23887"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=23887"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}